1.模数互质
#include<iostream>
#include<cstdio>
#include<climits>
#include<cstring>
#include<algorithm>
using namespace std;
int n,m[105],a[105],lcm=1;
int exgcd(int a,int b,int &x,int &y){//扩欧
if(!b){x=1,y=0;return a;}
int re=exgcd(b,a%b,x,y),tmp=x;
x=y,y=tmp-(a/b)*y;
return re;
}
int work(){
int i,j,d,x,y,re=0;
for(i=1;i<=n;i++)lcm=lcm*m[i];//因为互质所以直接这么写了
for(i=1;i<=n;i++){
int kl=lcm/m[i];
d=exgcd(kl,m[i],x,y);
x=(x%m[i]+m[i])%m[i];
re=(re+a[i]*x*kl)%lcm;
}
return re;
}
int main()
{
scanf("%d",&n);//组数
for(int i=1;i<=n;i++)scanf("%d%d",&m[i],&a[i]);//模数,余数
printf("%d",work());//原数
return 0;
}
2.模数不互质
#include<iostream>
#include<cstdio>
#include<climits>
#include<cstring>
#include<algorithm>
using namespace std;
#define LL long long
const int maxn=1e5+5;
int n;
LL exgcd(LL a,LL b,LL &x,LL &y){
if(!b){x=1,y=0;return a;}
LL re=exgcd(b,a%b,x,y),tmp=x;
x=y,y=tmp-(a/b)*y;
return re;
}
LL m[maxn],a[maxn];
LL work(){
LL M=m[1],A=a[1],t,d,x,y;int i;
for(i=2;i<=n;i++){
d=exgcd(M,m[i],x,y);//解方程
if((a[i]-A)%d)return -1;//无解
x*=(a[i]-A)/d,t=m[i]/d,x=(x%t+t)%t;//求x
A=M*x+A,M=M/d*m[i],A%=M;//日常膜一膜(划掉)模一模,防止爆
}
A=(A%M+M)%M;
return A;
}
int main()
{
int i,j;
while(scanf("%d",&n)!=EOF){
for(i=1;i<=n;i++)scanf("%lld%lld",&m[i],&a[i]);
printf("%lld\n",work());
}
return 0;
}