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#include#define isyeap(x) x%100!=0&&x%4==0||x%400==0 ?1:0//定义了一个宏 判断是否是闰年 int dayofmonth[13][2]={0,0,31,31,28,29,31,31,30,30,31,31,30,30,31,31,31,31,30,30,31,31,30,30,31,31};//定义了一个类,不仅可以用来表示日期,还能够自动计算出下一个日期 struct Date{ int day; int month; int year; void nextday(){ day++; if(day>dayofmonth[month][isyeap(year)]){ day=1; month++; if(month>12){ month=1; year++; } } }};int buf[5001][13][32];//返回绝对值 int abs(int x){ return x<0?-x:x;}int main(){ Date tmp; int cnt=0; tmp.day=1; tmp.month=1; tmp.year=0; //预处理数据 //先计算并保存每年每月每日与0年1月1日的日期差 while(tmp.year!=5001){ buf[tmp.year][tmp.month][tmp.day]=cnt; tmp.nextday(); cnt++; } int d1,m1,y1; int d2,m2,y2; while(scanf("%4d%2d%2d",&y1,&m1,&d1)!=EOF){ //使用%4d来读取该八位数输入的前四位并赋值给代表年的变量 scanf("%4d%2d%2d",&y2,&m2,&d2); printf("%d\n",abs(buf[y2][m2][d2]-buf[y1][m1][d1])+1); //hash的思想:用年、月 、日分别表示该数组下标 } return 0;}
这个解法真的很棒呀,简直是一气呵成~
那么当这个问题升级到英文版本,又回怎样呢?
题目大意为:输入一个日期 要求输入该天是星期几
那么我们只需要知道 1.今天是星期几 2.今天和所给的那天相隔几天
#include#include #define isyeap(x) x%100!=0&&x%4==0||x%400==0 ? 1:0int dayofmonth[13][2]={0,0,31,31,28,29,31,31,30,30,31,31,30,30,31,31,31,31,30,30,31,31,30,30,31,31};struct Date{ int day; int month; int year; void nextday(){ day++; if(day>dayofmonth[month][isyeap(year)]){ day=1; month++; if(month>12){ month=1; year++; } } }};int buf[3001][13][32];char monthname[13][20]={"","January","February","March","April","May","June","July","August","September","October","November","December"};char weeknAame[7][20]={"Sunday","Monday","Tuesady","Wednesday","Thursday","Friday","Saturday"};int main(){ Date tmp; int cnt=0; tmp.day=1; tmp.month=1; tmp.year=0; while(tmp.year!=3001){ buf[tmp.year][tmp.month][tmp.day]=cnt; tmp.nextday(); cnt++; } int d,m,y; char s[20]; while(scanf("%d%s%d",&d,&s,&y)!=EOF){ for(m=1;m<=12;m++){ if(strcmp(s,monthname[m])==0){ break; } } int days=buf[y][m][d]-buf[2019][4][10]; days+=3;//+3因为2019.4.10是星期三 puts(weekname[(days%7+7)%7]); //若为负数 则还要保证该下标为正 简单地处理成余数加7再求模 }return 0;}
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