SDUT 1941-Friday the Thirteenth(水)
Friday the Thirteenth
Time Limit: 1000ms Memory limit: 65536K 有疑问?点这里^_^
题目描写叙述
That is, does the 13th of the month land on a Friday less often than on any other day of the week? To answer this question, write a program that will compute the frequency that the 13th of each month lands on Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, and Saturday over a given period of N years. The time period to test will be from January 1, 1900 to December 31, 1900+N-1 for a given number of years, N. N is non-negative and will not exceed 400.
There are few facts you need to know before you can solve this problem:
- January 1, 1900 was on a Monday.
- Thirty days has September, April, June, and November, all the rest have 31 except for February which has 28 except in leap years when it has 29.
- Every year evenly divisible by 4 is a leap year (1992 = 4*498 so 1992 will be a leap year, but the year 1990 is not a leap year)
- The rule above does not hold for century years. Century years divisible by 400 are leap years, all other are not. Thus, the century years 1700, 1800, 1900 and 2100 are not leap years, but 2000 is a leap year.
Do not use any built-in date functions in your computer language.
Don't just precompute the answers, either, please.
PROGRAM NAME: friday
输入
One line with the integer N.
输出
For each input, there is an output correspond to it, each have the following format:
Seven space separated integers on one line. These integers represent the number of times the 13th falls on Saturday, Sunday, Monday, Tuesday, ..., Friday.
演示样例输入
20
演示样例输出
36 33 34 33 35 35 34
刷道存在感。。
题意 :问从1900年往后的n年里,每一个月的13号分布在星期1到7的情况。。这算哈希?反正暴力就能够了
#include <cstdio> #include <iostream> #include <algorithm> #include <cstring> #include <cctype> #include <cmath> #include <cstdlib> #include <vector> #include <queue> #include <set> #include <map> #include <list> #define ll long long using namespace std; const int INF=1<<27; const int maxn=1010; int n,num[8]; bool is_leap(int year) { if((year%4==0&&year%100!=0)||year%400==0) return 1; else return 0; } void solve() { memset(num,0,sizeof(num)); int xx=1; for(int year=1900;year<=1900+n-1;year++) { for(int month=1;month<=12;month++) { int tem; if(month==1||month==3||month==5||month==7||month==8||month==10||month==12) tem=31; else if(month==4||month==6||month==9||month==11) tem=30; else if(month==2&&is_leap(year)) tem=29; else if(month==2&&!is_leap(year)) tem=28; int tag=(xx+12)%7==0?7:(xx+12)%7; xx=(xx+tem)%7==0?7:(xx+tem)%7; num[tag]++; } } printf("%d %d",num[6],num[7]); for(int i=1;i<=5;i++) printf(" %d",num[i]); puts(""); } int main() { while(~scanf("%d",&n)) solve(); return 0; }