describe stdio.h ali ron max spa ces help rep

Strange Way to Express Integers

Time Limit: 1000MS		Memory Limit: 131072K
Total Submissions: 20640		Accepted: 6946

Description

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 a

₁, 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?

Input

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

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

2
8 7
11 9

Sample Output

31

Hint

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

題意 給出k個同余方程組：x mod ai = ri。求x的最小正值。如果不存在這樣的x，那麽輸出-1.不滿足所有的ai互質。

解析 https://blog.csdn.net/litble/article/details/75807726 中國擴展剩余定理

 1 #include<stdio.h>
 2 #define pb push_back
 3 #define mp make_pair
 4 #define fi first
 5 #define se second
 6 #define all(a) (a).begin(), (a).end()
 7 #define fillchar(a, x) memset(a, x, sizeof(a))
 8 #define huan printf("\n");
 9 #define debug(a,b) cout<<a<<" "<<b<<" "<<endl;
10 using namespace std;
11 const int maxn=1e5+10,inf=0x3f3f3f3f;
12 typedef long long ll;
13 ll chu[maxn],yu[maxn];
14 ll exgcd(ll a,ll b,ll &x,ll &y)   //&引用符號，修改x,y,函數返回的是a,b的最大公約數
15 {
16     if(b==0)
17     {
18         x=1;y=0;
19         return a;
20     }
21     ll gcd=exgcd(b,a%b,x,y);            //遞歸下去
22     ll temp=x;
23     x=y;y=temp-a/b*y;
24     return gcd;
25 }
26 ll CRT(int n)//中國剩余定理 除數互質
27 {
28     ll lcm=1,x,y,ans=0;
29     for(int i=0;i<n;i++)
30         lcm*=chu[i];   //除數互質
31     for(int i=0;i<n;i++)
32     {
33         ll temp=lcm/chu[i];
34         ll gcd=exgcd(temp,chu[i],x,y); //互質所以gcd肯定是1 x*temp%chu[i]=1 <=>x*temp+y*chu[i]=1
35         x=(x%chu[i]+chu[i])%chu[i];  //最小正整數解
36         ans=(ans+yu[i]*x*temp)%lcm;
37     }
38     return ans;
39 }
40 ll EXCRT(int n)//擴展中國剩余定理除數互不互質都可以
41 {
42     ll temp=chu[0],k=yu[0],x,y,t;
43     for(int i=1;i<n;i++)
44     {
45         ll gcd=exgcd(temp,chu[i],x,y);
46         if((yu[i]-k)%gcd)return -1;//無解
47         x*=(yu[i]-k)/gcd,t=chu[i]/gcd,x=(x%t+t)%t;
48         k=temp*x+k,temp=temp/gcd*chu[i],k%=temp;
49     }
50     k=(k%temp+temp)%temp;
51     return k;
52 }
53 int main()
54 {
55     int n;
56     while(scanf("%d",&n)!=EOF)
57     {
58         for(int i=0;i<n;i++)
59             scanf("%lld%lld",&chu[i],&yu[i]);
60         printf("%lld\n",CRT(n));
61     }
62     return 0;
63 }

poj 2891 中國剩余定理