zoukankan      html  css  js  c++  java
  • 连续取模-function

    2017-09-22 21:56:08

    The shorter, the simpler. With this problem, you should be convinced of this truth. 
       
      You are given an array AA of NN postive integers, and MM queries in the form (l,r)(l,r). A function F(l,r) (1lrN)F(l,r) (1≤l≤r≤N) is defined as: 
    F(l,r)={AlF(l,r1) modArl=r;l<r.F(l,r)={All=r;F(l,r−1) modArl<r. 
    You job is to calculate F(l,r)F(l,r), for each query (l,r)(l,r).

    InputThere are multiple test cases. 
       
      The first line of input contains a integer TT, indicating number of test cases, and TT test cases follow. 
       
      For each test case, the first line contains an integer N(1N100000)N(1≤N≤100000). 
      The second line contains NN space-separated positive integers: A1,,AN (0Ai109)A1,…,AN (0≤Ai≤109). 
      The third line contains an integer MM denoting the number of queries. 
      The following MM lines each contain two integers l,r (1lrN)l,r (1≤l≤r≤N), representing a query.

    OutputFor each query(l,r)(l,r), output F(l,r)F(l,r) on one line.Sample Input

    1
    3
    2 3 3
    1
    1 3

    Sample Output

    2

    代码如下:
    #include<iostream>
    #include<string.h>
    #include<stdlib.h>
    #include<stdio.h>
    
    using namespace std;
    
    #define MAXN 100010
    
    int a[MAXN],nex[MAXN];
    
    int main()
    {
        int t;
        scanf("%d",&t);
        while(t--)
        {
            int i,j,n,m;
            scanf("%d",&n);
            for(i = 1; i<=n; ++i)
            {
                scanf("%d",&a[i]);
            }
            for(i = 1;i<=n;++i)
            {
                nex[i] = -1;
                for(j = i+1;j<=n;++j)
                {
                    if(a[j]<=a[i])
                    {
                        nex[i] = j;
                        break;
                    }
                }
            }
            scanf("%d",&m);
            for(i = 0;i<m;++i)
            {
                int l,r;
                scanf("%d%d",&l,&r);
                int num = a[l];
                for(j = nex[l];j<=r;j = nex[j])
                {
                    if(j == -1)
                    {
                        break;
                    }
                    num%=a[j];
                }
                printf("%d
    ",num);
            }
        }
        return 0;
    }
  • 相关阅读:
    sql Test
    使用Team Foundation Server进行源代码管理
    幸运的秘密
    使用Dotmsn扩展Joymsg聊天机器人,使其同时支持QQ.MSN
    发布.net项目开发工具新版
    C#.Net项目生成器(ibatis)使用说明
    敏捷开发,Agile Development
    单元测试基础篇VS2008
    iBATIS.NET
    IBatis.Net学习笔记系列
  • 原文地址:https://www.cnblogs.com/pprp/p/7577351.html
Copyright © 2011-2022 走看看