USACO 1.1.3 Friday the Thirteenth

USACO 1.1.3   Friday the Thirteenth
                                     http://cerberus.delos.com:791/usacoprob2?a=owpOMMkldPW&S=friday
       题意：模拟。给定一个年数范围，计算在这些年中周一到周日为某个月13号的次数。
       ……不说了，早知道去做没做完的字符串题了。
       代码+细节：
 

      

#include<stdio.h>
/*
ID: 15257142
LANG: C
TASK: friday
*/

int main(){
int num,i,a[20],b,n,j,c;
FILE *fin  = fopen ("friday.in", "r");
FILE *fout = fopen ("friday.out", "w");
fscanf(fin,"%d",&num);
for(i = 0;i<=6;i++)
a[i] = 0;
b = 0;
for(i = 1900;i<=1900+num-1;i++){
c = 1;
b = b+13;
n = b%7;
if(n==0)
n = 7;
a[n-1]++;
b = b+18;

c = 2;
b = b+13;
n = b%7;
if(n==0)
n = 7;
a[n-1]++;
if((i%4==0&&i%100!=0)||(i%400==0))
b = b+16; /*闰年二月29天，普通为28天。这也能混。*/
else
b = b+15;

c = 3;
b = b+13;
n = b%7;
if(n==0)
n = 7;
a[n-1]++;
b = b+18;

c = 4;
b = b+13;
n = b%7; /*可以整除，正好是周日的情况。*/
if(n==0)
n = 7;
a[n-1]++;
b = b+17;

c =5;
b = b+13;
n = b%7;
if(n==0)
n = 7;
a[n-1]++;
b = b+18;

c = 6;
b = b+13;
n = b%7;
if(n==0)
n = 7;
a[n-1]++;
b = b+17;

c = 7;
b = b+13;
n = b%7;
if(n==0)
n = 7;
a[n-1]++;
b = b+18;

c = 8;
b = b+13;
n = b%7;
if(n==0)
n = 7;
a[n-1]++;
b = b+18;

c = 9;
b = b+13;
n = b%7;
if(n==0)
n = 7;
a[n-1]++;
b = b+17;

c = 10;
b = b+13;
n = b%7;
if(n==0)
n = 7;
a[n-1]++;
b = b+18;

c = 11;
b = b+13;
n = b%7;
if(n==0)
n = 7;
a[n-1]++;
b = b+17;

c = 12;
b = b+13;
n = b%7;
if(n==0)
n = 7;
a[n-1]++;
b = b+18;
}
fprintf(fout,"%d %d ",a[5],a[6]);
for(i = 0;i<=4;i++){
fprintf(fout,"%d",a[i]);
if(i!=4)
fprintf(fout," ");
}
fprintf(fout,"\n");
return 0;
}