题目是输入某年某月某日，判断这一天是这一年的第几天？
不改思路
#include<stdio.h>
int main(void)
{
    int x,y,z,n;
    printf("请输入年月日：");
    scanf("%d%d%d",&z,&y,&x);
    if(x>0&&y>0&&z>0&&x<31&&y<=12)
    {
        if((z%4==0&&z%100!=0)||(z%400==0))
        {
            if(y==1)
            {
                n=x+(y-1)*31;
                printf("是%d年的第%d天。\n",z,n);
            }
            else if(y==2&&x<28)
            {
                n=x+(y-1)*31;
                printf("是%d年的第%d天。\n",z,n);
            }
            else if(y%2==0)
            {
                if(x<=30)
                {
                    if(y<8)
                        n=x+y/2*31+28+(y/2-1)*30;
                    else if(y<12)
                        n=x+31*4+30*2+28+(y/2-3)*31+(y/2-4)*30;
                    printf("是%d年的第%d天。\n",z,n);
                }
                else
                    printf("你的输入有误！\n请从新执行程序！\n");
            }
            else
            {
                if(x<=31)
                {
                    if(y<8)
                        n=x+y/2*31+28+(y/2-1)*30;
                    else if(y<=12)
                        n=x+31*4+30*2+28+(y/2-3)*31+(y/2-4)*30;
                    printf("是%d年的第%d天。\n",z,n);
                }
                else
                    printf("你的输入有误！\n请从新执行程序！\n");
            }
        }
        else
        {
            if(y==1)
            {
                n=x+(y-1)*31;
                printf("是%d年的第%d天。\n",z,n);
            }
            else if(y==2&&x<29)
            {
                n=x+(y-1)*31;
                printf("是%d年的第%d天。\n",z,n);
            }
            else if(y%2==0)
            {
                if(x<=30)
                {
                    if(y<8)
                        n=x+y/2*31+29+(y/2-1)*30;
                    else if(y<12)
                        n=x+31*4+30*2+29+(y/2-3)*31+(y/2-4)*30;
                    printf("是%d年的第%d天。\n",z,n);
                }
                else
                    printf("你的输入有误！\n请从新执行程序！\n");
            }
            else
            {
                if(x<=31)
                {
                    if(y<8)
                        n=x+y/2*31+29+(y/2-1)*30;
                    else if(y<=12)
                        n=x+31*4+30*2+29+(y/2-3)*31+(y/2-4)*30;
                    printf("是%d年的第%d天。\n",z,n);
                }
                else
                    printf("你的输入有误！\n请从新执行程序！\n");
            }
        }
    }
    else
        printf("你的输入有误！\n请从新执行程序！\n");
    return 0;
}

麻烦顺便找一下漏洞

哇，这么多if -- else ,  那还不如把12个月分成switch语句算出天数来。
switch (month)
{
case 1:0 + day;
case 2: 31 + day
case 3: 31 + ? ( 平润年28/29)+ day;
……
case 12: 31 + ……+ 30 + day?
default：……
}

以前写的 ， 你参考下。
#include  <stdio.h>
int  main(void)
{
    int sum_day(int, int);
    int leap(int year);
    int exclude(int, int, int);
    int year, month, day, days;
    printf("Input date(year, month, day):");
    scanf("%d,%d,%d", &year, &month, &day);
    printf("%d/%d/%d", year, month, day);
    if (exclude(year, month, day))
    {
    days = sum_day(month, day);
    if (leap(year) && month >= 3)
        days = days + 1;
    printf("is the %dth day in the year.\n", days);
    }
    else
        printf("\nData error! Please enter again:\n") && main();
        
    return 0;
}

int sum_day(int month, int day)
{
    int  day_tab[13] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31} ;
    int i;
    for (i = 1; i < month; i++)
        day += day_tab[i];
    return(day);
}

int leap(int year)
{
    int leap;
    leap = year % 4 == 0 && year % 100 != 0 || year % 400 == 0 ;
    return(leap);
}

int exclude(int year, int month, int day)
{
    if ( (month <= 0 || month > 12 ) || (day <= 0 || day > 31))
        return (0);
    else
    {
        if (month == 2)
        {
            if((leap(year) && day > 29) || (!leap(year) && day > 28))
                return (0);
            else
                return (1);
        }
        else
          switch (month)
          {
            case 6:
            case 4:
            case 9:
            case 11: { if (day == 31) return (0);
                         else   return (1);
                         break;}
            default : return (1); break;
          }
    }
}

建立个数组就行了，月份及其天数，闰年就给2月加一天，然后就是几个月天数的累加了

我这个更简单！
#include<stdio.h>
#include<iostream>
using namespace std;
int yearday[2][13]={
    {0,31,28,31,30,31,30,31,31,30,31,30,31},
    {0,31,29,31,30,31,30,31,31,30,31,30,31}
} ;
int main()
{
    cout<<"输入年、月、日:"<<endl;
    int year,month,day,days=0;
    scanf("%d,%d,%d",&year,&month,&day);
    for(int i=0;i<month;i++)
       days+=yearday[year%4==0&&year%100||year%400==0][i];
    days+=day;
    cout<<year<<"年"<<month<<"月"<<day<<"日是这一年的第"<<days<<"天。"<<endl;
    return 0;
}

还有没有比这更简单的

#include<stdio.h>
int main()
{
    const int predays[] = {0, 0, 31, 59, 90, 120, 151, 181, 212, 243, 273, 304, 334 };
    int year, month, day;
    printf("输入年月日(以空格分隔):");
    scanf("%d %d %d", &year, &month, &day);
    printf("%d-%d-%d 是该年第 %d 天\n", year, month, day, predays[month] + day + (month > 2 ? !(year % 4) && year % 100 || !(year % 400) : 0));
    return 0;
}

这个貌似倒着写好些...

里面重复了那么多。

如果有非法输入怎么办？2000,2，29？