POJ-3061 前缀和+二分搜索 模板题

Subsequence

原题链接:http://poj.org/problem?id=3061

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

题解:一开始想用前缀和+双指针进行搜索,超时,巧用lower_bound()来搜索结果,真的很nice

 1 #include <string.h>
 2 #include <stdio.h>
 3 #include <algorithm>
 4 using namespace std;
 5 const int maxn=1e5+5;
 6 int B[maxn];
 7 int main(){
 8     int N;
 9     scanf("%d",&N);
10     int n,s,tmp=0;
11     while(N--){
12         memset(B,0,sizeof(B));
13         scanf("%d%d",&n,&s);
14         for(int i=1;i<=n;i++){
15             scanf("%d",&tmp);
16             if(i==1) B[i]=tmp;
17             B[i]=B[i-1]+tmp;
18         }
19         int j=0,summ=0,minL=maxn;
20         if (B[n]<s){
21             printf("%d
",0);
22             continue;
23         } 
24         if(B[1]>=s){
25             printf("%d
",1);
26             continue;
27         }
28         for(int i=1;B[i]+s<=B[n];i++){
29             j=lower_bound(B+i+1,B+n+1,B[i]+s)-B;
30             minL=min(minL,j-i);
31         }
32         printf("%d
",minL);
33     }
34     return 0;
35 }

菠萝写的更快地版本

 1 #include <string.h>
 2 #include <stdio.h>
 3 #include <algorithm>
 4 using namespace std;
 5 const int maxn=1e5+5;
 6 int nums[maxn];
 7 int main(){
 8     int N;
 9     scanf("%d",&N);
10     int n,s,tmp=0;
11     while(N--){
12         scanf("%d%d",&n,&s);
13         memset(nums,0,sizeof(nums));
14         for(int i=0;i<n;i++) scanf("%d",&nums[i]);
15         int sums=0,j=0,flag0=0,ans=0;
16         for (int i=0;i<n;i++){
17             int flag=1;
18             sums+=nums[i];
19             while (sums>=s){
20                 ans=i-j+1;
21                 sums-=nums[j];
22                 j+=1;
23                 flag=0;
24                 flag0=1;
25             }
26             if (flag && flag0){
27                 sums-=nums[j];
28                 j+=1;
29             }
30         }
31         printf("%d
", ans);
32     }
33     return 0;
34 }
原文地址:https://www.cnblogs.com/z-712/p/15079598.html