CRT用于求解一元线性同余方程组(模数互质),实际上模数不互质我们也可以解决,在之前的某篇文章里提过。如下
http://www.cnblogs.com/autsky-jadek/p/6596010.html
#include<cstdio>
using namespace std;
typedef long long ll;
ll m[4],a[4],D,ans;
void exgcd(ll a,ll b,ll &d,ll &x,ll &y){
if(!b){
d=a;
x=1;
y=0;
}
else{
exgcd(b,a%b,d,y,x);
y-=x*(a/b);
}
}
void CRT(int r){
ll M=1,Mi,x0,y0,d;
ans=0;
for(int i=1;i<=r;++i){
M*=m[i];
}
for(int i=1;i<=r;++i){
Mi=M/m[i];
exgcd(Mi,m[i],d,x0,y0);
ans=(ans+Mi*x0*a[i])%M;
}
if(ans<=D){
ans+=M;
}
}
int main(){
// freopen("poj1006.in","r",stdin);
int zu=0;
m[1]=23; m[2]=28; m[3]=33;
while(1){
++zu;
scanf("%lld%lld%lld%lld",&a[1],&a[2],&a[3],&D);
if(a[1]==-1 && a[2]==-1 && a[3]==-1 && D==-1){
break;
}
CRT(3);
printf("Case %d: the next triple peak occurs in %lld days.\n",zu,ans-D);
}
return 0;
}