https://icpcarchive.ecs.baylor.edu/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=4620

You are given an integer sequence of length N and another value X. You have to find a contiguous
subsequence of the given sequence such that the sum is greater or equal to X. And you have to find
that segment with minimal length.
Input
First line of the input file contains T the number of test cases. Each test case starts with a line
containing 2 integers N (1 ≤ N ≤ 500000) and X (−109 ≤ X ≤ 109
). Next line contains N integers
denoting the elements of the sequence. These integers will be between −109
to 109
inclusive.
Output
For each test case output the minimum length of the sub array whose sum is greater or equal to X. If
there is no such array, output ‘-1’.
Sample Input
3
5 4
1 2 1 2 1
6 -2
-5 -6 -7 -8 -9 -10
5 3
-1 1 1 1 -1
Sample Output
3
-1
3

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#include <vector>
#include <map>
#include <algorithm>
using namespace std; const int N = ;
const int INF = 0x3fffffff;
typedef long long LL;
#define met(a,b) (memset(a,b,sizeof(a))) struct node
{
LL x;
int Start;
} sum[N]; LL a[N]; int main()
{
int T;
scanf("%d", &T);
while(T--)
{
int n, i, Min=N, X, Start;
LL x; met(a, );
met(sum, ); scanf("%d%lld", &n, &x); for(i=; i<=n; i++)
scanf("%lld", &a[i]); for(i=; i<=n; i++)
{
if(sum[i-].x<= || i==)
{
sum[i].x = a[i];
sum[i].Start = i;
}
else
{
sum[i].x = sum[i-].x + a[i];
sum[i].Start = sum[i-].Start;
}
if(sum[i].x>=x)
{
Min = min(Min, i-sum[i].Start+);
X = sum[i].x, Start = sum[i].Start;
while(X>= && Start<=i)
{
X -= a[Start];
Start++;
if(X >= x)
{
sum[i].x = X;
sum[i].Start = Start;
Min = min(Min, i-Start+);
}
}
}
} printf("%d\n", Min!=N?Min:-);
} return ;
} /** 300
5 4
1 2 1 2 1
6 -2
-5 -6 -7 -8 -9 -10
5 3
-1 1 1 1 -1
8 6
1 1 1 1 1 2 3 4
6 5
4 -3 4 -1 2 2
6 6
-5 1 2 4 1 3
6 5
4 -3 4 -1 -2 2
6 5
-1 -1 -2 3 -2 5
4 5
3 -2 4 1
8 6
1 1 1 1 1 3 1 2
8 6
1 1 1 1 1 3 2 1
8 6
1 1 1 1 1 3 1 1 **/

6609 - Minimal Subarray Length的更多相关文章

  1. UVALive 6609 Minimal Subarray Length(RMQ-ST+二分)

    题意:给定长度为N的数组,求一段连续的元素之和大于等于K,并且让这段元素的长度最小,输出最小长度即可,若不存在这样的元素集合,则输出-1 题目链接:UVAlive 6609 做法:做一个前缀和pref ...

  2. UVALive 6609 Minimal Subarray Length (查找+构建排序数组)

    描述:给定n个整数元素,求出长度最小的一段连续元素,使得这段元素的和sum >= X. 对整个数组先求出sum[i],表示前i个元素的和,然后依次求出以a[i]为起点的,总和>= X的最小 ...

  3. UVA 12697 Minimal Subarray Length

    Minimal Subarray Length Time Limit: 3000ms Memory Limit: 131072KB This problem will be judged on UVA ...

  4. E - Minimal Subarray Length(连续区间和)

    题目链接 题意:给出n个数,求加和大于x的最短区间的区间长度. 如果前i个数字和为y,那么如果前j数字的和小于等于y-x,那么i-j就是一种可能的情况,我们对于所有的i找出前面最大的j就可以了,因为数 ...

  5. Individual Contest #1 and Private Training #1

    第一次的增补赛,也是第一场个人排位赛,讲道理打的和屎一样,手速题卡了好久还WA了好多发,难题又切不出来,这种情况是最尴尬的吧! Individual Contest #1: Ploblem D: 题意 ...

  6. Maximum Average Subarray II LT644

    Given an array consisting of n integers, find the contiguous subarray whose length is greater than o ...

  7. [leetcode]523. Continuous Subarray Sum连续子数组和(为K的倍数)

    Given a list of non-negative numbers and a target integer k, write a function to check if the array ...

  8. [poj1113][Wall] (水平序+graham算法 求凸包)

    Description Once upon a time there was a greedy King who ordered his chief Architect to build a wall ...

  9. *HDU 1392 计算几何

    Surround the Trees Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Other ...

随机推荐

  1. 模块二 hashlib模块、configparser模块、logging模块

    算法介绍 Python的hashlib提供了常见的摘要算法,如MD5,SHA1等等. 什么是摘要算法呢?摘要算法又称哈希算法.散列算法.它通过一个函数,把任意长度的数据转换为一个长度固定的数据串(通常 ...

  2. msyql [note] mysqld (mysqld 5.6.40) starting as process xxxx...

    my.ini有2个配置一个是客户端[client],一个是服务器端[mysqld] 如果把2个都设置为utf8,那么中文会显示乱码,或者无法插入中文 而把客户端设置为gbk,把服务器端设置为utf8就 ...

  3. encode/decode/bytes

    python3中如何将字符型转换成utf-8格式的bytes类型 str_me = '字符是我'.encode('utf-8') print(str_me) >>:b'\xe5\xad\x ...

  4. org.apache.commons.net.ftp

    org.apache.commons.NET.ftp Class FTPClient类FTPClient java.lang.Object Java.lang.Object继承 org.apache. ...

  5. andorid 多线程handler用法

    .xml <?xml version="1.0" encoding="utf-8"?> <LinearLayout xmlns:android ...

  6. mysql 优化之一

    提升速度 show  variables like 'innodb_flush_log_at_trx_commit'; 会显示为1 set global innodb_flush_log_at_trx ...

  7. Win7下Qt5的安装及使用

    1.安装Qt5 Qt5的安装比Qt4的安装简单多了,我装的是Qt5.4(qt-opensource-windows-x86-mingw491_opengl-5.4.0.exe),它集成了MinGW.Q ...

  8. HDU 6126.Give out candies 最小割

    Give out candies Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 524288/524288 K (Java/Other ...

  9. lua 2.2 变种

    1.修改 ~= 操作符为 != 2.取消 --[[ ]] 多行注释语法 下载源码

  10. python_docx制作word文档详细使用说明【转】

      目前网上对这一个库的介绍得很少,很零散,所以很多功能我是尽量参考其官网,但是官网上面很多功能目前只有说明文档,而代码并还没有及时更新,以至于按照官网上面做了,python却报错.比如:自定义表格的 ...