【POJ 2891】Strange Way to Express Integers（一元线性同余方程组求解）

Description

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Elina is reading a book written by Rujia Liu, which introduces a strange way to express non-negative integers. The way is described as following:
Choose k different positive integers a1, a2, …, ak. For some non-negative m, divide it by every ai (1 ≤ i ≤ k) to find the remainder ri. If a1, a2, …, ak are properly chosen, m can be determined, then the pairs (ai, ri) 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?

Input

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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 ai, ri (1 ≤ i ≤ k).

Output

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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.

Sample Input

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2
8 7
11 9

Sample Output

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31

Hint

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

Source

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POJ Monthly--2006.07.30, Static

参考代码

#include<queue>
#include<cmath>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#define ll long long
#define inf 1000000000
#define REP(i,x,n) for(int i=x;i<=n;i++)
#define DEP(i,x,n) for(int i=n;i>=x;i--)
#define mem(a,x) memset(a,x,sizeof(a))
using namespace std;
ll read(){
    ll x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-') f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
void Out(ll a){
    if(a<0) putchar('-'),a=-a;
    if(a>=10) Out(a/10);
    putchar(a%10+'0');
}
const int N=1e6+10;
void exgcd(ll a,ll b,ll &d,ll &x,ll &y){
     if(b==0){
         x=1;y=0;
         d=a;
     }else{
         exgcd(b,a%b,d,y,x),y-=x*(a/b);
     }
}
int a[N],r[N],n;
ll solve(){
    ll ta=a[1],tr=r[1],x,y,d;
    for(int i=2;i<=n;i++){
        exgcd(ta,a[i],d,x,y);
        if((r[i]-tr)%d) return -1;
        x=(r[i]-tr)/d*x%(a[i]/d);
        tr+=x*ta;ta=ta/d*a[i];
        tr%=ta;
     }
     return tr>0?tr:tr+ta;
}
int main(){
    while(~scanf("%d",&n)){
        REP(i,1,n) a[i]=read(),r[i]=read();
        Out(solve());
        puts("");
    }
    return 0;
}