Question

我需要减去由操作员输入的度数，分钟和秒数的2个角度。例如：30度50分45秒 - 10度20分53秒= 20度29分52秒。但我的代码返回32767分钟。

这是我的代码：

#include <stdio.h>

int main(int argc, char **argv)
{
    int g1,m1,s1; //timp 1
    int g2,m2,s2; //timp 2 
    int g,m,s; //calcule
    printf("Interval1:"); scanf("%d%d%d",&g1,&m1,&s1);
    printf("Interval2:"); scanf("%d%d%d",&g2,&m2,&s2);
    if(s1<s2) {
        s=(s1+60)-s2;
        m1=m1-1;
    } else if(m1<m2) {
        m=(m1-60)-m2;
        g1=g1-1;
    } else {
        s=s1-s2;
        m=m1-m2;
    }

    g=g1-g2;

    printf("%d %d %d",g,m,s);
    return 0;
}

回报是：

Interval1:30 50 45
Interval2:10 20 53
20 32767 52

------------------
(program exited with code: 0)
Press return to continue

Answer 1

这没关系？

#include <stdio.h>

int main(int argc, char **argv)
{
    int g1,m1,s1; //timp 1
    int g2,m2,s2; //timp 2 
    int g,m,s,mt,gt; //calcule
    printf("Interval1:"); scanf("%d%d%d",&g1,&m1,&s1);
    printf("Interval2:"); scanf("%d%d%d",&g2,&m2,&s2);
    if(s1<s2 && m1<m2) {
        s=(s1+60)-s2;
        mt=m1-1;
        m=mt-m2;
        g=g1-g2;
        m=(m1+60)-m2;
        gt=g1-1;
        g=gt-g2;
    } else if(s1<s2) {
        s=(s1+60)-s2;
        mt=m1-1;
        m=mt-m2;
        g=g1-g2;
    } else if(m1<m2) {
        m=(m1+60)-m2;
        gt=g1-1;
        g=gt-g2;
        s=s1-s2;
    } else {
        m=m1-m2;
        s=s1-s2;
        g=g1-g2;
    }

    printf("%d %d %d",g,m,s);
    return 0;
}

Answer 2

不那么复杂：

secs2gms(gms2secs(g1, m1, s1) - gms2secs(g2, m2, s2), &g, &m, &s);

...

long gms2secs(int g, int m, int s) {
    return g * 3600L + m * 60 + s;
}

void secs2gms(long secs, int *g, int *m, int *s) {
    *s = secs % 60;
    secs /= 60;
    *m = secs % 60;
    *g = secs / 60;
}

减去具有度，分和秒的2个角度

2 个答案: