zoukankan      html  css  js  c++  java
  • HDU——T 3579 Hello Kiki

    http://acm.hdu.edu.cn/showproblem.php?pid=3579

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 4208    Accepted Submission(s): 1617


    Problem Description
    One day I was shopping in the supermarket. There was a cashier counting coins seriously when a little kid running and singing "门前大桥下游过一群鸭,快来快来 数一数,二四六七八". And then the cashier put the counted coins back morosely and count again...
    Hello Kiki is such a lovely girl that she loves doing counting in a different way. For example, when she is counting X coins, she count them N times. Each time she divide the coins into several same sized groups and write down the group size Mi and the number of the remaining coins Ai on her note.
    One day Kiki's father found her note and he wanted to know how much coins Kiki was counting.
     
    Input
    The first line is T indicating the number of test cases.
    Each case contains N on the first line, Mi(1 <= i <= N) on the second line, and corresponding Ai(1 <= i <= N) on the third line.
    All numbers in the input and output are integers.
    1 <= T <= 100, 1 <= N <= 6, 1 <= Mi <= 50, 0 <= Ai < Mi
     
    Output
    For each case output the least positive integer X which Kiki was counting in the sample output format. If there is no solution then output -1.
     
    Sample Input
    2 2 14 57 5 56 5 19 54 40 24 80 11 2 36 20 76
     
    Sample Output
    Case 1: 341 Case 2: 5996
     
    Author
    digiter (Special Thanks echo)
     
    Source
     
    我的妈呀 CRT 还不互质
    shit
     1 #include <algorithm>
     2 #include <cstdio>
     3 
     4 using namespace std;
     5 
     6 int n,m[23],a[23];
     7 
     8 int exgcd(int a,int b,int &x,int &y)
     9 {
    10     if(!b)
    11     {
    12         x=1; y=0;
    13         return a;
    14     }
    15     int ret=exgcd(b,a%b,x,y),tmp=x;
    16     x=y; y=tmp-a/b*y;
    17     return ret;
    18 }
    19 int CRT(int m[],int a[])
    20 {
    21     int ret=a[1],mm=m[1];
    22     for(int i=2;i<=n;i++)
    23     {
    24         int gcd,x,y,b=m[i],tmp=a[i];
    25         int c=tmp-ret; gcd=exgcd(mm,b,x,y);
    26         if(c%gcd) return -1;
    27         x=x*c/gcd;
    28         int mod=b/gcd;
    29         x=(x%mod+mod)%mod;
    30         ret+=mm*x; mm*=mod;
    31     }
    32     if(!ret) ret+=mm;
    33     return ret;
    34 }
    35 
    36 int main()
    37 {
    38     int t; scanf("%d",&t);
    39     for(int k=1;k<=t;k++)
    40     {
    41         scanf("%d",&n);
    42         for(int i=1;i<=n;i++) scanf("%d",m+i);
    43         for(int i=1;i<=n;i++) scanf("%d",a+i);
    44         printf("Case %d: %d
    ",k,CRT(m,a));
    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。
  • 相关阅读:
    第一课 GCC入门
    第二课客户端链接Linux系统
    2014目标!!!!
    第一课Linux系统安装知识(2)
    android开发系列之ContentObserver
    android开发系列之数据存储
    android开发系列之视频断点续传
    稻盛和夫系列之活法一
    android开发系列之使用xml自定义控件
    android开发系列之MVP设计模式
  • 原文地址:https://www.cnblogs.com/Shy-key/p/7351362.html
Copyright © 2011-2022 走看看