数据连续化思想---判断一年中的第几天

普通代码实现（没用数据连续化思想）：

#include <stdio.h> int main(void) { int m; int year,month,day; int GetDay(int y,int m,int d); printf("please input year and month and day:"); scanf("%d %d %d",&year,&month,&day); m = GetDay(year,month,day); printf("Today is th year of %d day",m); } int GetDay(int y, int m, int d) //y年m月d日 { int s = 0; switch(m-1) { case 1: s=31; break; case 2: s=31+28; break; case 3: s=31+28+31; break; case 4: s=31+28+31+30; break; case 5: s=31+28+31+30+31; break; case 6: s=31+28+31+30+31+30; break; case 7: s=31+28+31+30+31+30+31; break; case 8: s=31+28+31+30+31+30+31+31; break; case 9: s=31+28+31+30+31+30+31+31+30; break; case 10: s=31+28+31+30+31+30+31+31+30+31; break; case 11: s=31+ 28+31+30+31+30+31+31+30+31+30; break; } if(m>2) //大于二月的话 看是否是闰年 如果是闰年 那要再加上一天 { if(y%100==0&&y%4!=0||y%400==0) { s++; } } return s+d; //返回这天是那一年的第几天 } 看看上面糟糕的代码，是不是感觉代码量太多了 现在来看看用数据连续化思想： #include <stdio.h> int main(void) { int m; int year,month,day; int GetDay(int y,int m,int d); printf("please input year and month and day:"); scanf("%d %d %d",&year,&month,&day); m = GetDay(year,month,day); printf("Today is th year of %d day",m); } int GetDay(int y,int m,int d) { int s = d; int List[12]={31,28,31,30,31,30,31,31,30,31,30,31};//把每个月比较特别的天数举出来 if(y%400==0 || (y%100 && y%4==0)) { ++List[1];//是的话2月多加一天 } for(y=0,--m;y<m;++y) { s+=List[y];//直接累加 } return s; }


大家比较下吧！