2750 - 端午节
时间限制 : 1 秒
内存限制 : 128 MB
2014年6月2日端午节，也是小粽子同学的18岁生日。可以推算出小粽子同学出生于星期日，经历了6575天（出生当天也算1天）。
题目输入
输入一个日期，格式是YYYY-MM-DD。如小粽子的生日是1996-06-02。
题目输出
<span style="font-size:14px;font-family:'Microsoft Yahei','serif';">2</span><span style="font-size:14px;font-family:宋体;">个数。第1个数表示此人从出生到</span><span style="font-size:14px;font-family:'Microsoft Yahei','serif';">2014</span><span style="font-size:14px;font-family:宋体;">年</span><span style="font-size:14px;font-family:'Microsoft Yahei','serif';">6</span><span style="font-size:14px;font-family:宋体;">月</span><span style="font-size:14px;font-family:'Microsoft Yahei','serif';">2</span><span style="font-size:14px;font-family:宋体;">日所经过的天数，第</span><span style="font-size:14px;font-family:'Microsoft Yahei','serif';">2</span><span style="font-size:14px;font-family:宋体;">个数表示这个人的出生日期是星期几。</span><span style="font-family:'Microsoft Yahei','serif';"></span>
输入/输出样例
输入格式
1975-09-01
输出格式
14155 1
C语言解答
#include<stdio.h> int main() { int year,month,day; //定义变量年、月、日 int total=153; //定义总的天数 int l,m,n,q,p; //定义中间变量 scanf("%d-%d-%d",&year,&month,&day); //输入数据 l=2014-year; m=l%4; n=l/4; switch(m) { case 1: total+=(1461*n); total+=365; switch(month) { case 1:total-=(day);break; case 2:total-=(31+day);break; case 3:total-=(59+day);break; case 4:total-=(90+day);break; case 5:total-=(120+day);break; case 6:total-=(151+day);break; case 7:total-=(181+day);break; case 8:total-=(212+day);break; case 9:total-=(243+day);break; case 10:total-=(273+day);break; case 11:total-=(304+day);break; case 12:total-=(334+day);break; } break; case 2: total+=(1461*n); total+=731; switch(month) { case 1:total-=(day);break; case 2:total-=(31+day);break; case 3:total-=(60+day);break; case 4:total-=(91+day);break; case 5:total-=(121+day);break; case 6:total-=(152+day);break; case 7:total-=(182+day);break; case 8:total-=(213+day);break; case 9:total-=(244+day);break; case 10:total-=(274+day);break; case 11:total-=(305+day);break; case 12:total-=(335+day);break; } break; case 3: total+=(1461*n); total+=1096; switch(month) { case 1:total-=(day);break; case 2:total-=(31+day);break; case 3:total-=(59+day);break; case 4:total-=(90+day);break; case 5:total-=(120+day);break; case 6:total-=(151+day);break; case 7:total-=(181+day);break; case 8:total-=(212+day);break; case 9:total-=(243+day);break; case 10:total-=(273+day);break; case 11:total-=(304+day);break; case 12:total-=(334+day);break; } break; case 0: total+=(1461*n); switch(month) { case 1:total-=(day);break; case 2:total-=(31+day);break; case 3:total-=(59+day);break; case 4:total-=(90+day);break; case 5:total-=(120+day);break; case 6:total-=(151+day);break; case 7:total-=(181+day);break; case 8:total-=(212+day);break; case 9:total-=(243+day);break; case 10:total-=(273+day);break; case 11:total-=(304+day);break; case 12:total-=(334+day);break; } break; } total+=1; q=total%7; switch(q) { case 6:p=3;break; case 5:p=4;break; case 4:p=5;break; case 3:p=6;break; case 2:p=7;break; case 1:p=1;break; case 0:p=2;break; } printf("%d\n%d",total,p); }
C++解答
#include<iostream> using namespace std; int main () { int y,m,d,c,s,w,n; char p,q; cin >> y >> p >> m >> q >> d; if (m==1) c=d; else if (m==2) c=31+d; else if (m==3) c=31+28+d; else if (m==4) c=31+28+31+d; else if (m==5) c=31+28+31+30+d; else if (m==6) c=31+28+31+30+31+d; else if (m==7) c=31+28+31+30+31+30+d; else if (m==8) c=31+28+31+30+31+30+31+d; else if (m==9) c=31+28+31+30+31+30+31+31+d; else if (m==10) c=31+28+31+30+31+30+31+31+30+d; else if (m==11) c=31+28+31+30+31+30+31+31+30+31+d; else if (m==12) c=31+28+31+30+31+30+31+31+30+31+30+d; if (m>2) { if (y%100==0) { if (y%400==0) c ++; } else if (y%4==0) c ++; } s = (y - 1) * 365 + ( y - 1) / 4 - (y - 1) / 100 + (y - 1) / 400 + c; n = (2014 - 1) * 365 + ( 2014 - 1) / 4 - (2014- 1) / 100 + (2014 - 1) / 400 + 154; cout << n-s<< endl; w = s % 7; switch (w) { case 0: cout << "7";break; case 1: cout << "1";break; case 2: cout << "2";break; case 3: cout << "3";break; case 4: cout << "4";break; case 5: cout << "5";break; case 6: cout << "6";break; } return 0; }