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  • POJ——T 2891 Strange Way to Express Integers

    http://poj.org/problem?id=2891

    Time Limit: 1000MS   Memory Limit: 131072K
    Total Submissions: 16849   Accepted: 5630

    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 a1a2…, ak. For some non-negative m, divide it by every ai (1 ≤ i ≤ k) to find the remainder ri. If a1a2, …, ak are properly chosen, m can be determined, then the pairs (airi) 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 airi (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.

    敲模板磨时间、、

     1 #include <algorithm>
     2 #include <cstdio>
     3 
     4 using namespace std;
     5 
     6 #define LL long long
     7 const int N(1005);
     8 LL n,a[N],m[N];
     9 
    10 LL exgcd(LL a,LL b,LL &x,LL &y)
    11 {
    12     if(!b)
    13     {
    14         x=1; y=0;
    15         return a;
    16     }
    17     LL ret=exgcd(b,a%b,x,y),tmp=x;
    18     x=y; y=tmp-a/b*y;
    19     return ret; 
    20 }
    21 LL CRT()
    22 {
    23     LL ret=m[1],tot=a[1];
    24     for(int i=2;i<=n;i++)
    25     {
    26         LL x,y,tt=a[i],c=m[i]-ret;
    27         LL gcd=exgcd(tot,tt,x,y);
    28         if(c%gcd) return -1;
    29         x=x*c/gcd;
    30         LL mod=tt/gcd;
    31         x=(x%mod+mod)%mod;
    32         ret+=x*tot; tot*=mod;
    33     }
    34     if(!ret) ret+=tot;
    35     return ret;
    36 }
    37 
    38 int main()
    39 {
    40     for(;~scanf("%lld",&n);)
    41     {
    42         for(int i=1;i<=n;i++)
    43             scanf("%lld%lld",a+i,m+i);
    44         printf("%lld
    ",CRT());
    45     }
    46     return 0;
    47 }
    ——每当你想要放弃的时候,就想想是为了什么才一路坚持到现在。
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  • 原文地址:https://www.cnblogs.com/Shy-key/p/7352845.html
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