POJ-1006(中国剩余定理)

http://poj.org/problem?id=1006

中国剩余定理：

先求出

X1%23==1,X1%28==0,X1%33==0;

X2%28==1,X2%23==0,X2%33==0;

X3%33==1,X3%28==0,X3%23==0;

代码如下：

#include<stdio.h>
int main()
{
int i;
for(i=1;i<100;++i)
if((28*33)*i%23==1)
break;
printf("%d\n",(28*33)*i);
for(i=1;i<100;++i)
if((23*33)*i%28==1)
break;
printf("%d\n",(23*33)*i);
for(i=1;i<100;++i)
if((23*28)*i%33==1)
break;
printf("%d\n",(23*28)*i);
while(1);
return 0;
}

得：X1=5544， X2=14421，X3=1288

故：所求天数为 N=5544×p+14421*e+1288*i-d;
代码如下：

#include<stdio.h>
int main()
{
int p,e,i,d,k=0;
while(scanf("%d%d%d%d",&p,&e,&i,&d),!(p==-1&&e==-1&&i==-1&&d==-1))
{
k++;
int m=(5544*p+14421*e+1288*i-d+21252)%21252;
if(m<=0)
printf("Case %d: the next triple peak occurs in %d days.\n",k,21252+m);
else
printf("Case %d: the next triple peak occurs in %d days.\n",k,m);
}
return 0;
}