zoukankan      html  css  js  c++  java
  • B

    B - Subsequence

    Time Limit: 2000/1000 MS (Java/Others)      Memory Limit: 128000/64000 KB (Java/Others)
    Submit Status

    Problem Description

    A sequence of N positive integers (10 < N < 100 000), each of them less than or equal 10000, and a positive integer S (S < 100 000 000) are given. Write a program to find the minimal length of the subsequence of consecutive elements of the sequence, the sum of which is greater than or equal to S.

    Input

    The first line is the number of test cases. For each test case the program has to read the numbers N and S, separated by an interval, from the first line. The numbers of the sequence are given in the second line of the test case, separated by intervals. The input will finish with the end of file.

    Output

    For each the case the program has to print the result on separate line of the output file.if no answer, print 0.

    Sample Input

    2
    10 15
    5 1 3 5 10 7 4 9 2 8
    5 11
    1 2 3 4 5

    Sample Output

    2
    3
     1 #include <stdio.h>
     2 #include <queue>
     3 #include <cstring>
     4 #include <cstdlib>
     5 #include <algorithm>
     6 #include <vector>
     7 #include <map>
     8 #include <set>
     9 #include <ctime>
    10 #include <cmath>
    11 #include <cctype>
    12 using namespace std;
    13 typedef long long LL;
    14 const int N=1e5+10;
    15 const int INF=0x3f3f3f3f;
    16 int cas=1,T;
    17 int n,a[N],s;
    18 int main()
    19 {
    20 //    freopen("1.in","w",stdout);
    21 //    freopen("1.in","r",stdin);
    22 //    freopen("1.out","w",stdout);
    23     scanf("%d",&T);
    24     while(T--)
    25     {
    26         scanf("%d%d",&n,&s);
    27         for(int i=0;i<n;i++) scanf("%d",a+i);
    28         int ans=INF;
    29         for(int i=0,j=0,sum=0;i<n;i++)
    30         {
    31             while(sum<s && j<n)
    32             {
    33                 sum+=a[j];
    34                 j++;
    35             }
    36             if(sum>=s) ans=min(ans,j-i);
    37             sum-=a[i];
    38         }
    39         if(ans==INF) ans=0;
    40         printf("%d
    ",ans);
    41     }
    42 //    printf("time=%.3lf
    ",(double)clock()/CLOCKS_PER_SEC);
    43     return 0;
    44 }
    solve.cpp

    原题:POJ3061

    题解:

    二分或滑动窗口,题解是滑动窗口
    二分:枚举每个位置,然后二分找出刚好区间和大于S的另一个位置,记录最小长度
    滑动窗口:维护两个下标动态记录长度

  • 相关阅读:
    YunTable开发日记(16)教程(0.9版RC)
    DevOps,不是一个传说!
    C/C++可变参数函数(转载) zjhfqq的专栏 52RD博客_52RD.com
    c++大写
    fabric install depernedecy
    mysql 在int ,bit类型中都 支持not 取反操作
    关于realloc的原理,与实现方法 C/C++ / C语言
    了解YunTable | 人云亦云
    如何让编译时的出错提示由中文变为英文的? 查看主题 • Ubuntu中文论坛
    Tomcat配置技巧Top 10
  • 原文地址:https://www.cnblogs.com/cdyboke/p/7010491.html
Copyright © 2011-2022 走看看