#include<stdio.h>
main()
{
int h1, min1, sec1, a;
int h2, min2, sec2, b;
char choice;

printf("A for add\n");
printf("S for subtract\n");
printf("Please enter A or S only:");
scanf("%c", &choice);

switch (choice)
{
    case 'A':    printf("\nPlease enter the first clock in correct Military format:\n");
                 scanf("%d%d%d", &h1, &min1, &sec1);
                 printf("Clock1=%d seconds\n", a=h1*3600+min1*60+sec1);
                 printf("\nPlease enter the second clock in correct Military format:\n");
                 scanf("%d%d%d", &h2, &min2, &sec2);
                 printf("Clock2=%d seconds\n", b=h2*3600+min2*60+sec2);
                 if (a+b==86400)
                 {
                 printf("\nThen Clock3 is 00:00:00\n");
                 }
                 else
                 {
                     if (a+b>86400)
                     {
                     printf("\nThen Clock3 is %d:%d:%d hrs\n", (a+b)/3600-24, (a+b)%3600/60, (a+b)%60);
                     }
                      else
                     {
                      printf("\nThen Clock3 is %d:%d:%d hrs\n", (a+b)/3600, (a+b)%3600/60, (a+b)%60);
                     }
                 }
                 break;
     
     case 'S':   printf("\nPlease enter the first clock in correct Military format:\n");
                 scanf("%d%d%d", &h1, &min1, &sec1);
                 printf("Clock1=%d seconds\n", a=h1*3600+min1*60+sec1);
                 printf("\nPlease enter the second clock in correct Military format:\n");
                 scanf("%d%d%d", &h2, &min2, &sec2);
                 printf("Clock2=%d seconds\n", b=h2*3600+min2*60+sec2);
                 if (a-b==0)
                 {
                 printf("\nThen Clock3 is 00:00:00\n");
                 }
                 else
                    if (a-b<0)
                    {
                     printf("\nThen Clock3 is %d:%d:%dhrs\n", ( a+24*3600-b)/3600, (a+24*3600-b)%3600/60, (a+24*3600-b)%60);
                    }
                    else
                    {
                    printf("\nThen Clock3 is %d:%d:%dhrs\n", ( a-b)/3600, (a-b)%3600/60, (a-b)%60);
                    }
                 break;
                 default: printf("\nInvalid choice, please try again.\n\n");
}
                  
}

外面套个循环（while（1）），在循环末位询问，如果选择结束，break就行了

具体语句怎么写呢，我试了好几种语句都不行。

[ 本帖最后由 winglesswu 于 2013-2-5 12:24 编辑 ]

#include<stdio.h>
main()
{
    int h1, min1, sec1, a;
    int h2, min2, sec2, b;
    char choice;

    while(1)
    {
        printf("A for add\n");
        printf("S for subtract\n");
        printf("Please enter A or S only:");
        scanf("%c", &choice);

        switch (choice)
        {
        case 'A':
            printf("\nPlease enter the first clock in correct Military format:\n");
            scanf("%d%d%d", &h1, &min1, &sec1);
            printf("Clock1=%d seconds\n", a=h1*3600+min1*60+sec1);
            printf("\nPlease enter the second clock in correct Military format:\n");
            scanf("%d%d%d", &h2, &min2, &sec2);
            printf("Clock2=%d seconds\n", b=h2*3600+min2*60+sec2);
            if (a+b==86400)
            {
                printf("\nThen Clock3 is 00:00:00\n");
            }
            else
            {
                if (a+b>86400)
                {
                    printf("\nThen Clock3 is %d:%d:%d hrs\n", (a+b)/3600-24, (a+b)%3600/60, (a+b)%60);
                }
                else
                {
                    printf("\nThen Clock3 is %d:%d:%d hrs\n", (a+b)/3600, (a+b)%3600/60, (a+b)%60);
                }
            }
            break;

        case 'S':
            printf("\nPlease enter the first clock in correct Military format:\n");
            scanf("%d%d%d", &h1, &min1, &sec1);
            printf("Clock1=%d seconds\n", a=h1*3600+min1*60+sec1);
            printf("\nPlease enter the second clock in correct Military format:\n");
            scanf("%d%d%d", &h2, &min2, &sec2);
            printf("Clock2=%d seconds\n", b=h2*3600+min2*60+sec2);
            if (a-b==0)
            {
                printf("\nThen Clock3 is 00:00:00\n");
            }
            else if (a-b<0)
            {
                printf("\nThen Clock3 is %d:%d:%dhrs\n", ( a+24*3600-b)/3600, (a+24*3600-b)%3600/60, (a+24*3600-b)%60);
            }
            else
            {
                printf("\nThen Clock3 is %d:%d:%dhrs\n", ( a-b)/3600, (a-b)%3600/60, (a-b)%60);
            }
            break;
        default:
            printf("\nInvalid choice, please try again.\n\n");
        }
        getchar();
        printf("Again?(y/n)");
        if(getchar()=='n')
            break;
        fflush(stdin);
    }

}
OK的话给分，过两天给大家贺年啊

[ 本帖最后由 dsl975708035 于 2013-2-5 13:22 编辑 ]

#include<stdio.h>
main()
{
    int h1, min1, sec1, a;
    int h2, min2, sec2, b;
    char choice;

    while(1)
    {
        printf("A for add\n");
        printf("S for subtract\n");
        printf("R for end\n");
        printf("Please enter A or S only:");
        scanf("%c", &choice);

        switch (choice)
        {
        case 'R':return;
        case 'A':
            printf("\nPlease enter the first clock in correct Military format:\n");
            scanf("%d%d%d", &h1, &min1, &sec1);
            printf("Clock1=%d seconds\n", a=h1*3600+min1*60+sec1);
            printf("\nPlease enter the second clock in correct Military format:\n");
            scanf("%d%d%d", &h2, &min2, &sec2);
            printf("Clock2=%d seconds\n", b=h2*3600+min2*60+sec2);
            if (a+b==86400)
            {
                printf("\nThen Clock3 is 00:00:00\n");
            }
            else
            {
                if (a+b>86400)
                {
                    printf("\nThen Clock3 is %d:%d:%d hrs\n", (a+b)/3600-24, (a+b)%3600/60, (a+b)%60);
                }
                else
                {
                    printf("\nThen Clock3 is %d:%d:%d hrs\n", (a+b)/3600, (a+b)%3600/60, (a+b)%60);
                }
            }
            break;

        case 'S':
            printf("\nPlease enter the first clock in correct Military format:\n");
            scanf("%d%d%d", &h1, &min1, &sec1);
            printf("Clock1=%d seconds\n", a=h1*3600+min1*60+sec1);
            printf("\nPlease enter the second clock in correct Military format:\n");
            scanf("%d%d%d", &h2, &min2, &sec2);
            printf("Clock2=%d seconds\n", b=h2*3600+min2*60+sec2);
            if (a-b==0)
            {
                printf("\nThen Clock3 is 00:00:00\n");
            }
            else if (a-b<0)
            {
                printf("\nThen Clock3 is %d:%d:%dhrs\n", ( a+24*3600-b)/3600, (a+24*3600-b)%3600/60, (a+24*3600-b)%60);
            }
            else
            {
                printf("\nThen Clock3 is %d:%d:%dhrs\n", ( a-b)/3600, (a-b)%3600/60, (a-b)%60);
            }
            break;
        default:
            printf("\nInvalid choice, please try again.\n\n");
        }
        getchar();
        printf("Again?(y/n)");
        if(getchar()=='n')
            break;
        fflush(stdin);
    }

}

谢谢两位的解答，但是输入第一次的时候没问题，当选择y重新输入时显示：

Then Clock3 is 11:9:59hrs
Again?(y/n)y
A for add
S for subtract
Please enter A or S only:A
Again?(y/n)A for add
S for subtract
Please enter A or S only:A

这个Again选择又出现了。

我测试没问题啊

谢谢心灯甚亮，请问可以把具体的嵌套写出来吗。下面是我写的：
#include<stdio.h>
main()
{
    int h1, min1, sec1, a;
    int h2, min2, sec2, b;
    char yn, choice;

    while(1)
    {
        printf("This is a program of adding or substracting time which results in another time. Have fun.\n");
        printf("\nPlease follow the instructions. Thank you!\n");
        printf("A for add\n");
        printf("S for subtract\n");
        printf("Please enter A or S only:");
        scanf("%c", &choice);

        switch (choice)
        {
        case 'A':
            printf("\nPlease enter the first clock in correct Military format:\n");
            scanf("%d%d%d", &h1, &min1, &sec1);
            printf("Clock1=%d seconds\n", a=h1*3600+min1*60+sec1);
            printf("\nPlease enter the second clock in correct Military format:\n");
            scanf("%d%d%d", &h2, &min2, &sec2);
            printf("Clock2=%d seconds\n", b=h2*3600+min2*60+sec2);
            if (a+b==86400)
            {
                printf("\nThen Clock3 is 00:00:00\n");
            }
            else
            {
                if (a+b>86400)
                {
                    printf("\nThen Clock3 is %d:%d:%d hrs\n", (a+b)/3600-24, (a+b)%3600/60, (a+b)%60);
                }
                else
                {
                    printf("\nThen Clock3 is %d:%d:%d hrs\n", (a+b)/3600, (a+b)%3600/60, (a+b)%60);
                }
            }
            break;

        case 'S':
            printf("\nPlease enter the first clock in correct Military format:\n");
            scanf("%d%d%d", &h1, &min1, &sec1);
            printf("Clock1=%d seconds\n", a=h1*3600+min1*60+sec1);
            printf("\nPlease enter the second clock in correct Military format:\n");
            scanf("%d%d%d", &h2, &min2, &sec2);
            printf("Clock2=%d seconds\n", b=h2*3600+min2*60+sec2);
            if (a-b==0)
            {
                printf("\nThen Clock3 is 00:00:00\n");
            }
            else if (a-b<0)
            {
                printf("\nThen Clock3 is %d:%d:%dhrs\n", ( a+24*3600-b)/3600, (a+24*3600-b)%3600/60, (a+24*3600-b)%60);
            }
            else
            {
                printf("\nThen Clock3 is %d:%d:%dhrs\n", ( a-b)/3600, (a-b)%3600/60, (a-b)%60);
            }
            break;
        }
        getchar();
        printf("Again?(y/n)");
        if(getchar()=='n')
            break;
        fflush(stdin);
    }

}
但是具体运行的时候会出现下面的结果：
Again?(y/n)y
This is a program of adding or substracting time which results in another time. Have fun.

Please follow the instructions. Thank you!
A for add
S for subtract
Please enter A or S only:S
Again?(y/n)This is a program of adding or substracting time which results in another time. Have fun.

Please follow the instructions. Thank you!
A for add
S for subtract
Please enter A or S only:
问题是我选了y时，出现：
This is a program of adding or substracting time which results in another time. Have fun.

Please follow the instructions. Thank you!
A for add
S for subtract
Please enter A or S only:
这时候我再选择S时，则再次出现：
Again?(y/n)This is a program of adding or substracting time which results in another time. Have fun.

Please follow the instructions. Thank you!
A for add
S for subtract
Please enter A or S only:
下面选择就没问题了。

again(y/n)
这里你直接按了回车吧。。。
getchar()接受了"\n",你可以改成按除’n‘意外的任意键继续，n退出

[ 本帖最后由 心灯甚亮 于 2013-2-6 12:08 编辑 ]

您说的对，按了回车键就等于选择了y。那么有没有办法输入y继续循环，而不用回车键呢？谢谢。