T2000

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#include <stdio.h>

int main()
{
	char ch[3], temp;
	int i, j;
	while((scanf("%s", ch)) != EOF)
	{
		for(j = 0; j < 2; j++)
		{
			for(i = 0; i < 2 - j; i++)
			{
				if(ch[i] > ch[i + 1])
					temp = ch[i];
					ch[i] = ch[i + 1];
					ch[i + 1] = temp;
				}
			}
		}
		for(i = 0; i < 3; i++)
		{
			printf("%c", ch[i]);
			if(ch[i + 1] != '\0')
				printf(" ");
			else printf("\n");
		}
	}
	return 0;
}

T2001

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#include <stdio.h>
#include <math.h>           //数学库，使用sqrt()
#define LEN 20

int main()
{
	double set[4];
	double d[LEN];
	int i = 0, j;
	while((scanf("%lf %lf %lf %lf", &set[0], &set[1], &set[2], &set[3])) == 4)
    {
    	d[i] = sqrt((set[2] - set[0])*(set[2] - set[0])+ (set[3] - set[1])*(set[3] - set[1]));
    	i++;
    }
    j = i;
    for(i = 0; i < j; i++)
    {
        printf("%.2lf\n", d[i]);
    }
    return 0;
}

T2002

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#include <stdio.h>
#define PI 3.1415927
#define LEN 20

int main()
{
	double r;
	double s[LEN];
	int i = 0, j;
	const double k = 4.00/3.00;
	while((scanf("%lf", &r)) == 1)
	{
		s[i] = k*PI*r*r*r;
		i++;
	}
	j = i;
	for(i = 0; i < j; i++)
	{
		printf("%.3lf\n", s[i]);
	}
	return 0;
}

T2003

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#include <stdio.h>
#define LEN 20

int main()
{
	double num, abs[LEN];
	int i = 0, j;
	while((scanf("%lf", &num)) == 1)
	{
		if(num >= 0)
			abs[i] = num;
		else abs[i] = -num;
		i++;
	}
	j = i;
	for(i = 0; i < j; i++)
	{
		printf("%.2lf\n", abs[i]);
	}
	return 0;
}

T2004

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#include <stdio.h>
#define F "Score is error!"
#define LEN 20

int main()
{
	int score, i = 0, j = 0, temp, false[LEN];
	char rank[LEN];
	while((scanf("%d", &score)) == 1)
	{
		if(score >= 90 && score <= 100)
			rank[i] = 'A';
		else if(score >= 80 && score <= 89)
			rank[i] = 'B';
		else if(score >= 70 && score <= 79)
			rank[i] = 'C';
		else if(score >= 60 && score <= 69)
			rank[i] = 'D';
		else if(score >= 0 && score <= 59)
			rank[i] = 'E';
		else {
			false[j] = i;
			j++;                    //false[]记录错误值
		}
		i++;
	}
	temp = i;
	for(i = 0, j = 0; i < temp; i++)
	{
		if( i != false[j])
			printf("%c\n", rank[i]);
		else{
			printf("%s\n", F);
			j++;
		}
	}
	return 0;
}

T2005

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#include<stdio.h>
int main() {
	int day[12] = {31,0,31,30,31,30,31,31,30,31,30,31};	
	int year, month, date;
	int total = 0;

	while(scanf("%d/%d/%d", &year, &month, &date) != EOF) {
		if((year % 400 == 0) || ((year % 100 != 0) && (year % 4 == 0))) 		{		
			day[1] = 29;									
		} else {
			day[1] = 28;	
		}//输入技巧：scanf中包含'/',不用字符串读取，避免类型转换
	 
		int i = 0;		//月份计数
		while(i < month - 1) {
			total +=  day[i];
			i++;
		}
	 
		total += date;
		printf("%d\n",total);
		total = 0;
	}
	return 0;

}

T2006

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#include <stdio.h>
#define ROW 20

int main()
{
	int n, num, i, j = 0;
	int mul[ROW];
	while((scanf("%d", &n)) != EOF )
	{
	    mul[j] = 1;
		for(i = 0; i < n; i++)
		{
			scanf("%d", &num);
			if(num % 2 == 1)
                mul[j] *= num;
        }
		j++;
	}
	i = j;
	for(j = 0; j < i; j++)
	{
		printf("%d\n", mul[j]);
	}
	return 0;

}

T2007

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#include <stdio.h>
#define LEN 20

int main()
{
	int min, max, temp;
	int i, j = 0;
	long long int sum1[LEN], sum2[LEN];
	while((scanf("%d %d", &min, &max)) != EOF)
	{
		sum1[j] = sum2[j] = 0;
		if(min > max)//一定要记得大小比较！！！
        {
            temp = min;
            min = max;
            max = temp;
        }
		for(i = min; i <= max; i++)
		{
			if(i % 2 == 1)
				sum1[j] += i * i * i;
			else if(i % 2 == 0)
				sum2[j] += i * i;
		}
		j++;
	}
	for(i = 0; i < j; i++)
	{
		printf("%lld %lld\n", sum2[i], sum1[i]);
	}
	return 0;
}

T2008

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#include <stdio.h>
#define LEN 20

int main()
{
	int  n, i, j = 0;
	double a[LEN];
	int c1[LEN] = {0}, c2[LEN] = {0}, c3[LEN] = {0};
	while(1)
	{
	    scanf("%d", &n);
	    if(n == 0)
            break;
		for(i = 0; i < n; i++){
			scanf("%lf", &a[i]);
			if(a[i] == 0)
				c1[j]++;
			if(a[i] < 0)
				c2[j]++;
			else if(a[i] > 0)
                c3[j]++;
		}
		j++;
	}
	for(i = 0; i < j; i++){
	printf("%d %d %d\n", c2[i], c1[i], c3[i]);
	}
	return 0;
}

T2009

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#include <stdio.h>
#include <math.h>
#define LEN 20

int main()
{
	double sum[LEN] = {0};
	double n;
	int m;
	int i, j = 0;
	while((scanf("%lf %d", &n, &m)) != EOF)
	{
		for(i = 0; i < m; i++)
		{
			sum[j] +=  n;
			n = sqrt(n);
		}
		j++;
	}
	for(i = 0; i < j; i++)
		printf("%.2lf\n", sum[i]);
	return 0;
}

T2010

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#include <stdio.h>
#include <stdlib.h>

int main()
{
    int m, n, temp;
    int i, g, s, b;
    int count;
    while((scanf("%d %d", &m, &n)) != EOF)
    {
        count = 0;                          //每次清零计数器
        if(m > n){
            temp = m;
            m = n;
            n = temp;
        }
        for(i = m; i <= n; i++)
        {
            g = i % 10;
            s = (i / 10) % 10;
            b = i / 100;                    //计算个、十、百位
            if(i == b*b*b + s*s*s + g*g*g){
                if(count >= 1)
                    printf(" ");
                count++;
                printf("%d", i);
            }
        }
        if(count == 0)
            printf("no\n");
        else printf("\n");
    }

    return 0;

}

T2011

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#include<stdio.h>
#define LEN 100

int main()
{
	int m, n[LEN], i, j, k = 0;
	double result[LEN] = {0};
	scanf("%d", &m);
	for(i = 0; i < m; i++)
	{
		scanf("%d", &n[i]);
		for(j = 1; j <= n[i]; j++){
			if(j % 2 == 1)
				result[k] += (double)(1.00/j);
			else result[k] -= (double)(1.00/j);
		}
		k++;
	}
	j = k;
	for(i = 0; i < j; i++)
		printf("%.2lf\n", result[i]);
	return 0;
}

T2012

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#include<stdio.h>

int main()
{
    int n, m, j, x, y, t;
    while(scanf("%d %d", &x, &y) != EOF)
    {
        if(x == 0 && y == 0)
            break;
        else{
            for(n = x; n <= y; n++){                   
                m = n * n + n + 41;
                for(j = 2; j < m; j++){                 //判断素数
                    if(m % j != 0)
                        t = 1;
                    else{
                        t = 0;
                        break;
                    }
                }
                if(t == 0){
                    printf("Sorry\n");
                    break;
                }
                else
                    continue;
            }
            if(t != 0)
                printf("OK\n");
        }
    }
    return 0;
}

T2013

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#include <stdio.h>

int main()
{
    int n, i;
    int sum;
    while((scanf("%d", &n)) != EOF)
    {
        sum = 1;
        for(i = 1; i < n; i++)
        {
            sum = (sum + 1) * 2;
        }
        printf("%d\n", sum);
    }
    return 0;

}

T2014

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#include <stdio.h>
int main ()
{
    int n,i,j;
    float a[100],x;
    while(scanf("%d",&n)!=EOF)
    {
        for(i=0;i<n;i++)
            scanf("%f",&a[i]);
        for(i=0;i<n;i++)       
        {
            for(j=i+1;j<n;j++)
            {
                if(a[i]>a[j])
                    x=a[i],a[i]=a[j],a[j]=x;
            }
        }
        x=0;
        for(i=1;i<n-1;i++)   
        {
            x=x+a[i];
        }
        x=x/(n-2);
        printf("%.2f\n",x);
    }
    return 0;
}

T2015

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#include <stdio.h>

int main()
{
	int n, m, i;
	int sum;
	int av;
	while((scanf("%d %d", &n, &m)) != EOF)
	{
		i = 0;
		sum = 0;
		for(i = 2; i <= 2 * n; i = i + 2)
		{
			sum += i;
			if(((i / 2) % m) == 0)
			{
				av = sum / m;
				sum = 0;
				printf("%d", av);
				if(n % m != 0)
                    printf(" ");          //处理输出的形式，这里容易presentation error
                if(n % m == 0)
                {
                    if(i != 2 * n)
                        printf(" ");
                    else printf("\n");   //处理完后光标跳至下一行等待输入
                }
            }

		}
		sum = 0;
		av = 0;
		if(n % m != 0)                   //对最后不足m项的处理
		{
			for(i = 2 * (n - (n % m)) + 2; i <= 2 * n; i = i + 2)
			{
				sum += i;
			}
			printf("%d\n", sum/(n % m));
		}
	}
	return 0;

}

T2016

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#include<stdio.h>

int main()
{
    int n;
    while((scanf("%d", &n)) != EOF)
    {
        if(n == 0)                   //结束条件
            break;
        int m[100];
        int min = 0;
        for(int i = 0; i < n; i++)
        {
            scanf("%d", &m[i]);
            if(m[min] > m[i])            //记录下标，方便直接交换
                min = i;
        }
        int temp;
        temp = m[min];
        m[min] = m[0];
        m[0] = temp;
        for(int i = 0; i < n - 1; i++)
            printf("%d ", m[i]);              //防止presentation error
        printf("%d\n", m[n - 1]);
    }
    return 0;
}

T2017

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#include <stdio.h>
#include <string.h>            //使用strlen()

int main()
{
	char ch[100];
	int n;
	int count;
	scanf("%d", &n);
	for(int i = 1; i <= n; i++)
	{
		count = 0;
		scanf("%s", ch);
		for(int j = 0;  j < strlen(ch); j++)
		{
			if(ch[j] <= '9' && ch[j] >= '0')     //数字对应ASCII码编号是自身，但要带''
				count++;
		}
			printf("%d\n", count);
	}
	return 0;

}

T2018

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#include <stdio.h>

int fbnq4(int n);         //声明函数，仿斐波那契

int main()
{
	int n;
	int count;
	while((scanf("%d", &n)) != EOF)          //主函数作为fbnq4()的驱动函数
    {
    	if(n == 0)
    		break;
    	count = fbnq4(n);
    	printf("%d\n", count);
    }
    return 0;
}

int fbnq4(int n)
{
	if(n <= 4)
	 return n;
	 else return fbnq4(n - 1) + fbnq4(n - 3);   //仿斐波那契递归
}

T2019

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#include <stdio.h>

int main()
{
	int a[100];
	int n, m;
	while((scanf("%d %d", &n, &m)) != EOF)
	{
		if(n == 0 && m == 0)
			break;
		for(int i = 0; i < n; i++){
			scanf("%d", &a[i]);
		}
		for(int i = 0; i < n; i++)
		{
			if(a[i] < m)
				continue;
			else if(a[i] > m){
				for(int j = n - 1; j >= i; j--)      //每个值后移，同时数组扩大
					a[j + 1] = a[j];
				a[i] = m;
				for(int i = 0; i < n + 1; i++){
					if(i != n)
					printf("%d ", a[i]);
					else printf("%d\n", a[n]);
				}
				break;             //直接break外层循环
			}
		}
	}
	return 0;
}

T2020

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#include <stdio.h>
#include <stdlib.h>

int main()
{
    int n, a[100];
    while(scanf("%d", &n) != EOF)
    {
        if(n == 0)  break;

        for(int i = 0;i < n; i++)
        {
            scanf("%d", &a[i]);
        }
        
        for(int i = 0; i < n-1; i++)
            for(int j = 0; j < n-i-1; j++)
            {
                if(abs(a[j]) < abs(a[j+1]))
                {
                    int temp = a[j];
                    a[j] = a[j+1];
                    a[j+1] = temp;
                }
            }

        for(int i = 0; i < n; i++)
        {
            if(i == n-1) {
            	printf("%d\n", a[i]);
                break;
            }
            printf("%d ", a[i]);
        }
    }

    return 0;
}

T2021

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#include<stdio.h>
//典型贪心算法，从最大面额开始（话说给的示例工资也太少吧喂！ 
int count(int n){                 
	int sum = 0;
	int a[6] = {100,50,10,5,2,1};     //人民币面值种类 
	for(int i = 0; i<6; i++){
		sum += n/a[i];                //人民币张数计算 
		n = n%a[i];
	} 
	return sum;
} 
int main() 
{
    int i,j,n;
    int a[100];
    while(scanf("%d",&n) != EOF) {
    	int k = 0;                  //更新k的值 
    	if(n == 0)return 0;
        for(i = 0; i < n; i++){
        	scanf("%d", &a[i]);
		}    
		for(j = 0; j < n; j++){
		    k += count(a[j]);	
		}
		printf("%d\n", k);
    }
    return 0;

}

T2022

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#include <stdio.h>
#include <stdlib.h>

//题干太长了吧！！！
int main()
{
    int m, n;
    int point[100][100];        //二维数组自带行与列
    while((scanf("%d %d", &m, &n)) != EOF)
    {
        int max = 0, x = 0, y = 0;
        for(int i = 0; i < m; i++)            //两层循环处理输入
        {
            for(int j = 0; j < n; j++){
                scanf("%d", &point[i][j]);
                if(abs(max) < abs(point[i][j])){
                    max = point[i][j];
                    x = i + 1;             //记录行
                    y = j + 1;             //记录列
                }
            }
        }
        printf("%d %d %d\n", x, y, max);
    }
}

T2023

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#include <stdio.h>
#include <stdlib.h>

int main()
{
    int n, m;
    int a[50][5];
    int count;
    int is_big = 0;
    while((scanf("%d %d", &n, &m)) != EOF)
    {
        int i ,j, tot;
        double b[5] = {0}; //有必要记录下每科的平均分
        count = 0;
        for(i = 0; i < n; i++)
        {
            for(j = 0; j < m; j++)
                scanf("%d", &a[i][j]);
        }
        for(i = 0; i < n; i++)          //第一行输出
        {
            tot = 0;
            for(j = 0; j < m; j++)
            {
                tot += a[i][j];
            }
            if(i == n - 1)
                printf("%.2lf\n", (double)tot/(double)m);
            else printf("%.2lf ", (double)tot/(double)m);  //被presentation error整怕了
        }
        for(j = 0; j < m; j++)
        {
            tot = 0;
            for(i = 0; i < n; i++)
            {
                tot += a[i][j];
            }
            b[j] = (double)tot/(double)n;    //需要记录的b的数量与科目数相同
            if(j == m - 1)
                printf("%.2lf\n", b[j]);
            else printf("%.2lf ", b[j]);
        }
        for(i = 0; i < n; i++)
        {
            for(j = 0; j < m; j++)
            {
                if(a[i][j] >= b[j])
                    is_big = 1;               //引入is_big判断
                else is_big = 0;
            }
            if(is_big)
                count++;
        }
        printf("%d\n", count);
        printf("\n");
    }
    return 0;
}

T2024

好家伙，帮我复习了一下用的不多的两个函数

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#include<stdio.h>
#include<ctype.h>

int main()
{
	int a, b, i;
	char ch[50];
	while (scanf("%d", &a) != EOF)
	{
		getchar();
		while (a--)
		{
			is_legal = 0;
			gets(ch);
			if (ch[0] != '_' && isalpha(ch[0]) == 0)
			{
				printf("no\n");
				continue;
			}
			for (i = 1; ch[i] != '\0'; i++)
			{
				if (ch[i] != '_' && isalpha(ch[i]) == 0 && isalnum(ch[i]) == 0)
				{
					printf("no\n");
					is_legal = 1;
					break;
				}
			}
			if (is_legal != 1)
				printf("yes\n");
		}
	}
	return 0；
}

T2025

本来用strchr()写了一个版本：

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#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int main()
{
    char ch[100];
    while(gets(ch))
    {
        char max = ch[0];
        int idx[100];
        for(int i = 1; i < strlen(ch); i++)
        {
            if((int)max < (int)ch[i])
                max = ch[i];
        }
        char *p = strchr(ch, max);
        int i = 0;
        while(p != NULL)
        {
            idx[i] = p - ch;
            p = strchr(p + 1, max);
            i++;
        }
        int j;
        for(i = 0, j = 0; i < strlen(ch); i++)
        {
            if(i != idx[j])
                printf("%c", ch[i]);
            else{
                printf("%c(max)", ch[i]);
                j++;
            }
        }
        printf("\n");
    }
    return 0;
}

yysy，我把所有可能的输入都试遍了，都没有问题，但就是过不了oj，没办法写了一般版本：(不过此题用字符串函数是杀鸡焉用牛刀)

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#include<stdio.h>

int main(void)
{
	char a[101];
	int i, j, k;
	while (gets(a))
	{
		j = 0;
		k = 0;
		for (i = 0; a[i] != '\0'; i++)    //两种标记循环结束的方法
		{
			if (a[i] > j)
			{
				j = a[i];
				k = i;
			}
		}
		for (i = 0; a[i] != '\0'; i++)
		{
			if (a[i] == a[k])
				printf("%c(max)", a[i]);
			else
				printf("%c", a[i]);
		}
		printf("\n");
	}
	return 0;
}

T2026

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#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int main()
{
    char sen[100];
    while(gets(sen))
    {
        int i;
        sen[0] -= 32;
        for(i = 0; i < strlen(sen); i++)
        {
            if(sen[i] == ' ')
                sen[i+1] -= 32;
        }
        puts(sen);
    }
    return 0;
}
//也可以用另一种解法，即include一个ctype.h使用toupper()直接转换大写

T2027

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#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int main()
{
    int n, j = 0;
    char s[100];
    scanf("%d", &n);
    getchar();           //！一定要有，不然'\n'会被gets吃到
    while(++j <= n)
    {
        gets(s);
        int c1 = 0, c2 = 0, c3 = 0, c4 = 0, c5 = 0;
        for(int i = 0; i < strlen(s); i++)
        {
            if(s[i] == 'a')
                c1++;
            if(s[i] == 'e')
                c2++;
            if(s[i] == 'u')
                c3++;
            if(s[i] == 'i')
                c4++;
            else if(s[i] == 'o')
                c5++;
        }
        if(j <= n - 1)
            printf("a:%d\ne:%d\ni:%d\no:%d\nu:%d\n\n", c1, c2, c4, c5, c3);
        if(j == n)
             printf("a:%d\ne:%d\ni:%d\no:%d\nu:%d\n", c1, c2, c4, c5, c3);
    }
    return 0;
}
//注意PE！搞心态orz
//当然switch更简洁

T2028

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#include <stdio.h>
#include <stdlib.h>

int main()
{
    int n, is_times  = 0;
    while((scanf("%d", &n)) != EOF)
    {
        int a[100];
        for(int i = 0; i < n; i++)
        {
            scanf("%d", &a[i]);
        }
        int j = 1;
        while(j++)               //枚举法找最大公倍数，时间复杂度为o(n)
        {
            for(int i = 0; i < n; i++)          
            {
                if(j%a[i] == 0)
                    is_times = 1;
                else {
                    is_times = 0;
                    break;
                }
            }
            if(is_times){
                printf("%d\n", j);
                break;
            }
        }
    }
    return 0;
}

T2029

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#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int main()
{
    int n;
    char s[100];
    scanf("%d", &n);
    getchar();
    int count = 0, is_huiwen = 0;
    while(count++ < n && gets(s))
    {
        for(int i = 0, j = strlen(s) - 1; i < strlen(s) / 2 ; i++, j--)
        {
            if(s[i] == s[j])
                is_huiwen = 1;
            else {
                is_huiwen = 0;
                break;
        }
        }
            if(is_huiwen == 1)
                printf("yes\n");
            else if(is_huiwen == 0)
                printf("no\n");
    }
    return 0;

}

T2030

**汉字机内码

//在做这题之前我也曾想过计算机如何储存汉字，但没有深究过，感谢2030题给我这个机会🌹🌹

GBK编码中，汉字占两个字节（两个char类型），且两个char类型对应编码由十六进制转换为十进制均为负数，若想用printf打印出汉字，必需要两个%c相连，否则就不知道会打印什么东西了。如：

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char *str = "你好，世界！"
char *chr = str;
while(*chr != '\0')
{
	if(*chr < 0)
	{
		printf("%c%c", *chr, *(chr+1));
		chr += 2;
	}
}
......

那么此题思路就很简单了：

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#include <stdio.h>
#include <string.h>

int main()
{
    int n;
    scanf("%d", &n);
    getchar();
    char ch[10000];
    while(n--)
    {
        gets(ch);
        int count = 0;
        for(int i = 0; i < strlen(ch); i++)
        {
            if(ch[i] < 0)
                count++;
        }
        printf("%d\n", count / 2);           //根据两个字节对应一个汉字的特性
    }
    return 0;
}

T2031

复习一下十进制转R进制原理：

十进制数除以R取余并记录余数，最后将每步得到的余数倒置即可。

T2032

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#include <stdio.h>
#include <stdlib.h>

int main()
{
    int n;
    int tri[30][30];

    while((scanf("%d", &n)) != EOF)
    {
        tri[0][0] = 1;
        tri[0][1] = 0;
        for(int i = 0; i < n; i++)
        {
            for(int j = 0; j <= i; j++)
            {
                if(j == 0 && i != 0)
                {
                    tri[i][j] = 1;
                    printf("%d ", tri[i][j]);
                }
                if(j == 0 && i == 0)
                    printf("%d", tri[i][j]);
                if(j == i && i != 0)
                {
                    tri[i][j] = 1;
                    printf("%d", tri[i][j]);
                }
                else if(j != 0 && j != i){
                    tri[i][j] = tri[i-1][j] + tri[i-1][j-1] ;
                    printf("%d ", tri[i][j]);
                }
            }
            printf("\n");
        }
        printf("\n");
    }

    return 0;
}

调PR调了半天/恼，每行后面不让多出来一个空格orz

后来发现其实可以在给每行末尾添加一个0值、

网上的一个没有用二维数组的版本：

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#include <stdio.h>
#include <string.h>
int main()
{
	int i, j, n;
	int yh[32];

	while(scanf("%d",&n) != EOF)
	{
		memset(yh, 0, sizeof(yh));//初始化
		yh[0]=1;
		for(i = 0; i < n; i++)
		{
			printf("%d", 1);
			for(j = i; j; j--)
				yh[j] += yh[j-1];
			for(j = 1; j <= i; j++)
				printf(" %d", yh[j]);//注意空格
			putchar('\n');
		}
		putchar('\n');
	}
	return 0;
}

相当于每次输出一行后，该一维数组的值即为该行，然后每个位置自身加上前面一个数即可得到新行。

T2033

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#include <stdio.h>
#include <string.h>

int main()
{
    int n;
    scanf("%d", &n);
    int ah, am, as, bh, bm, bs, rh, rm, rs;
    int flag = 0;        //是否进位的标记，可以共用
    while(n--)
    {
        scanf("%d %d %d %d %d %d", &ah, &am, &as, &bh, &bm, &bs);
        if((bs + as) < 60){
            rs = bs + as;
            flag = 0;
        }
        else {
                rs = bs + as - 60;
                flag = 1;
        }
        if((bm + am) < 60 && flag == 0){
            rm = bm + am;
            flag = 0;
        }
        if((bm + am) < 60 && flag == 1){
            rm = bm + am + 1;
            flag = 0;
        }
        if((bm + am) > 60 && flag == 0){
            rm = bm + am - 60;
            flag = 1;
        }
        if((bm + am) > 60 && flag == 1){
            rm = bm + am - 60 + 1;
            flag = 1;
        }
        if(flag)
            rh = ah + bh + 1;
        else rh = ah + bh;
        printf("%d %d %d\n", rh, rm, rs);
    }
    return 0;

}