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:
    605. Can Place Flowers
    1184. Distance Between Bus Stops
    1711. Count Good Meals
    1710. Maximum Units on a Truck
    566. Reshape the Matrix
    980. Unique Paths III
    212. Word Search II
    每日总结
    每日总结
    每日总结
  • 原文地址:https://www.cnblogs.com/pprp/p/7577351.html
Copyright © 2011-2022 走看看