POJ2891StrangeWaytoExpressIntegers

Choose k different positive integers a₁, a₂, …, a_k. For some non-negative m, divide it by every a_i (1 ≤ i ≤ k) to find the remainder r_i. If a₁, a₂, …, a_k are properly chosen, m can be determined, then the pairs (a_i, r_i) can be used to express m.

“It is easy to calculate the pairs from m, ” said Elina. “But how can I find m from the pairs?”

Since Elina is new to programming, this problem is too difficult for her. Can you help her?

The input contains multiple test cases. Each test cases consists of some lines.

• Line 1: Contains the integer k.
• Lines 2 ~ k + 1: Each contains a pair of integers a_i, r_i (1 ≤ i ≤ k).

Output the non-negative integer m on a separate line for each test case. If there are multiple possible values, output the smallest one. If there are no possible values, output -1.

All integers in the input and the output are non-negative and can be represented by 64-bit integral types.

题解：中国剩余定理非互素版本

代码：

#include
#include
#include
#define LL long long
#define maxn 100009
using namespace std;

int n;
LL m[maxn],r[maxn];

LL exgcd(LL a,LL b,LL &x,LL &y){
    if(b==0){
        x=1;y=0;
        return a;
    }
    LL t,r=exgcd(b,a%b,x,y);
    t=x;x=y;y=t-a/b*y;
    return r;
}

LL CRT(LL m[],LL r[],LL n){
    LL x,y,gcd,M=m[1],R=r[1];
    for(int i=2;i<=n;i++){
        LL gcd=exgcd(M,m[i],x,y);
        if((r[i]-R)%gcd)return -1;
        x=(r[i]-R)/gcd*x%(m[i]/gcd);
        R+=x*M;//新的余数R=r1+x*m[1] 
        M=M/gcd*m[i];//新的M，lcm(m1,m2) 
        R%=M;
    }
    return R>0?R:R+M;
}

int main(){
    while(~scanf("%d",&n)){
        for(int i=1;i<=n;i++)
         scanf("%lld%lld",&m[i],&r[i]);
        printf("%lld\n",CRT(m,r,n));
    }
    return 0;
}