zoukankan      html  css  js  c++  java
  • uvalive2678Subsequence

    题意:有n个正整数组成一个序列,给定整数S,求长度最短的连续序列,使得他们的和大于等于S

    分析:设输入的序列为A[i], i=1..n, 构造前缀数组B[j], j=1..n, B[j]=B[j-1]+A[j], 规定B[0]=0, 当B[j]-B[i-1]>=s的时候i增加,直至B[j]-B[i]<s, 然后更新最短的满足条件的序列的长度j-i+1,复杂度为O(n)

    代码:

    View Code
     1 #include <stdio.h>
     2 #include <iostream>
     3 using namespace std;
     4 #define DEBUG
     5 const int MAXN = 100000 + 10;
     6 int a[MAXN];
     7 int b[MAXN];
     8 int min(int a, int b){
     9     return a<b?a:b;
    10 }
    11 int main(){
    12 #ifndef DEBUG
    13     freopen("in.txt", "r", stdin);
    14 #endif
    15     int n, s;
    16     while(scanf("%d%d", &n, &s)!=EOF){
    17         int i;
    18         b[0]=0;
    19         for(i=1; i<=n; i++){
    20             scanf("%d", &a[i]);
    21             b[i] = b[i-1] + a[i];
    22         }
    23         int ans=n+1;
    24         i=1;
    25         int j;
    26         for(j=1; j<=n; j++){
    27             if(b[i-1]>b[j]-s) continue;
    28             while(b[i]<=b[j]-s) i++;
    29             ans = min(ans, j-i+1);
    30         }
    31         if(ans==n+1) ans=0;
    32         printf("%d\n", ans);
    33     }
    34     return 0;
    35 }

    如果暴力,O(n3)显然不行;如果用upper_bound则为O(n*log(n))

    Greatness is never a given, it must be earned.
  • 相关阅读:
    Go
    Go
    Go
    Go
    Go
    Go
    爬虫常用相关库
    Go
    python基础第7天(day19)
    python基础第五天(day17)元组,集合,字符串操作 字符编码:
  • 原文地址:https://www.cnblogs.com/zjutzz/p/2910056.html
Copyright © 2011-2022 走看看