计算出任意二个时间之间的天数，不知道哪里出错了 - C语言论坛

#include<stdio.h> int a[13]={0,31,28,31,30,31,30,31,31,30,31,30,31}; int ye2(int n,int x,int y) { int i,sum=0; if(x>=2) { if(n%4==0) { sum=sum+1; for(i=x-1;i>0;i++) { sum=sum+a[i]; } }else{ for(i=x-1;i>0;i--) { sum=sum+a[i]; } } sum=sum+y; } return sum; } int ye1(int n,int x,int y) { int i,sum; if(x<=2) { if(n%4==0) { sum+=1; for(i+1;i<13;i++) { sum+=a[i]; } }else{ for(i=x+1;i<13;i++) { sum+=a[i]; } } sum=sum+y; } return sum; } int nian(int x,int y) { int i,sum=0; for(i=x+1;i<y;i++) { if(i%4==0) { sum+=366; }else{ sum+=365; } } return sum; } int main() { int sum,n1,y1,r1,n2,y2,r2; printf("请输入第一个日期(xxxx-xx-xx):"); scanf("%d-%d-%d",&n1,&y1,&r1); printf("请输入第二个日期(xxxx-xx-xx):"); scanf("%d-%d-%d",&n2,&y2,&r2); if(n1=n2) { sum=ye1(n1,y1,r1)+ye2(n2,y2,r2); }else{ sum=nian(n1,n2)+ye1(n1,y1,r1)+ye2(n2,y2,r2); } printf("第一个日期到第二个日期一共%d天\n",sum); return 0; }

第 3 楼

得分:7

你的逻辑错了，请先想清楚日期间隔具体是怎么算的。

程序代码：

#include <stdio.h>

int a[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};

//日期是此年的第几天
int ye(int n,int x,int y)
{
        int i,sum=0;
        if(x>=3)  //x>=3
        {
                if((n%400==0)||(n%100!=0&&n%4==0))  //for leap year
                {
                    sum=1;
                }

                for(i=x-1;i>0;i--)
                {
                    sum=sum+a[i];
                }

                sum=sum+y;
        }
        else //x<=2
        {
            if(x==1)  sum=y;
            else  sum=31+y;
        }

        return sum;
}
//年与年之前的日期差，仅以1月1日算
int nian(int x,int y)
{
        int i,sum=0,tmp;
        if(x>y)  //make sure x>y
        {
            tmp=x;
            x=y;
            y=tmp;
        }

        for(i=x;i<y;i++)
        {
                if((i%400==0)||(i%100!=0&&i%4==0))
                {
                        sum+=366;
                }else{
                        sum+=365;
                }
        }

        return sum;
}

int main()
{
        int sum,n1,y1,r1,n2,y2,r2;
        printf("The first date:(xxxx-xx-xx):");
        scanf("%d-%d-%d",&n1,&y1,&r1);
        printf("The second date(xxxx-xx-xx):");
        scanf("%d-%d-%d",&n2,&y2,&r2);
        if(n1==n2)  //n1==n2
        {
                sum=ye(n1,y1,r1)-ye(n2,y2,r2);  //
                if(sum<0)  sum=(-1)*sum;  //
        }
        else
        {   
                //不同年日期的间隔天数=年与年之前的日期差，仅以1月1日算+大年的当前天数-小年的当前天数
                sum=nian(n1,n2);
                if(n1>n2)  sum=sum+ye(n1,y1,r1)-ye(n2,y2,r2);
                else  sum=sum+ye(n2,y2,r2)-ye(n1,y1,r1);

        }

        printf("There are %d days long.\n",sum);

        return 0;
}

[此贴子已经被作者于2016-4-15 09:04编辑过]

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#include<stdio.h> int a[13] = { 0,31,28,31,30,31,30,31,31,30,31,30,31 }; int ye2(int n, int x, int y) { int i, sum = 0; if (x >= 2) { if (n % 4 == 0) { sum = sum + 1; for (i = x - 1; i>0; i--) //应该是i-- { sum = sum + a[i]; } } else { for (i = x - 1; i>0; i--) { sum = sum + a[i]; } } sum = sum + y; } return sum; } int ye1(int n, int x, int y) { int i, sum = 0;//sum没有初始化就使用了 if (x <= 2) { if (n % 4 == 0) { sum += 1; for (i = 1; i<13; i++)//应该是i = 1不是i+ 1 { sum += a[i]; } } else { for (i = x + 1; i<13; i++) { sum += a[i]; } } sum = sum + y; } return sum; } int nian(int x, int y) { int i, sum = 0; for (i = x + 1; i<y; i++) { if (i % 4 == 0) { sum += 366; } else { sum += 365; } } return sum; } int main() { int sum, n1, y1, r1, n2, y2, r2; printf("请输入第一个日期(xxxx-xx-xx):"); scanf("%d-%d-%d", &n1, &y1, &r1); printf("请输入第二个日期(xxxx-xx-xx):"); scanf("%d-%d-%d", &n2, &y2, &r2); if (n1 = n2) { sum = ye1(n1, y1, r1) + ye2(n2, y2, r2); } else { sum = nian(n1, n2) + ye1(n1, y1, r1) + ye2(n2, y2, r2); } printf("第一个日期到第二个日期一共%d天\n", sum); return 0; }