zoukankan      html  css  js  c++  java
  • Sequence(尺取)

    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
    11
    1 2 3 4 5

    Sample Output

    2
    3
    题解:求连续子序列和大于等于s的最小长度。可以尺取,若sum<s,左指针右移,若sum>=s右指针左移,当sum<s时左指针继续右移,满足条件打擂台即可。
     1 #include<cstring>
     2 #include<cstdio>
     3 #include<algorithm>
     4 #define Inf 0x3f3f3f3f
     5 #define  ll   long long
     6 using namespace std;
     7 ll str[123400];
     8 int main()
     9 {
    10     ll i,j,m,n,t;
    11     scanf("%lld",&t);
    12     while(t--)
    13     {
    14         scanf("%lld%lld",&m,&n);
    15         for(i=0;i<m;i++)
    16         {
    17             scanf("%lld",&str[i]);
    18         }
    19         ll sum=0,cnt=0,cur=0,flag=0,ans=Inf,count1=0;
    20         while(cur<m)
    21         {
    22             while(count1<m&&sum<=n)
    23             {
    24                 sum+=str[count1++];
    25             }
    26             if(sum<n)
    27                 break;
    28                 flag=1;
    29             sum-=str[cur];
    30             ans=min(ans,count1-cur);
    31             cur++;
    32         }
    33         if(flag)
    34         printf("%lld
    ",ans);
    35         else
    36             puts("0");
    37     }
    38     return 0;
    39 
    40 }
  • 相关阅读:
    Sublime Text 3 支持的热门插件推荐
    Sublime text 2/3 中 Package Control 的安装与使用方法
    UML类图中箭头和线条的含义和用法
    this guy gonna be a daddy
    PHP设计模式之:单例模式
    PHP FTP操作类( 上传、拷贝、移动、删除文件/创建目录 )
    php memcache 基础操作
    获取IP地址方法
    短信发送
    使用Shell脚本对Linux系统和进程资源进行监控
  • 原文地址:https://www.cnblogs.com/moomcake/p/9343255.html
Copyright © 2011-2022 走看看