【题目】
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
Many test cases will be given. 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 there isn't such a subsequence, print 0 on a line by itself.
Sample Input
10 15 5 1 3 5 10 7 4 9 2 8 5 11 1 2 3 4 5
Sample Output
2 3
【分析】
本题最直接的思路是二重循环，枚举子序列的起点和终点。代码如下（输入数据已存入数组A[1]~A[n]）。
int ans = n+1;
for(int i = 1; i <= n; i++)
for(int j = i; j <= n; j++) {
int sum = 0;
for(int k = i; k <= j; k++) sum += A[k];
if(sum >= S) ans = min(ans, j-i+1);
}
printf("%d\n", ans == n+1 ? 0 : ans);
很可惜，上述程序的时间复杂度是O(n3)的，因此，当n达到100 000的规模后，程序将无能为力。有一个方法可以降低时间复杂度，即常见的前缀和技巧。令Bi=A1+A2+…+Ai，规定B0=0，则可以在O(1)时间内求出子序列的值：Ai+Ai+1+…+Aj=Bj-Bi-1。这样，时间复杂度降为O(n2)，代码如下。
B[0] = 0;
for(int i = 1; i <= n; i++) B[i] = B[i-1] + A[i];
int ans = n+1;
for(int i = 1; i <= n; i++)
for(int j = i; j <= n; j++)
if(B[j] - B[i-1] >= S) ans = min(ans, j-i+1);
printf("%d\n", ans == n+1 ? 0 : ans);
遗憾的是，本题的数据规模太大，O(n2)时间复杂度的算法也太慢。不难发现，只要同时枚举起点和终点，时间复杂度不可能比O(n2)更低，所以必须另谋他路。比如，是否可以不枚举终点，只枚举起点，或者不枚举起点，只枚举终点呢？
我们首先试试只枚举终点。对于终点j，我们的目标是要找到一个让Bj-Bi-1≥S，且i尽量大（i越大，序列长度j-i+1就越小）的i值，也就是找一个让Bi-1≤Bj-S最大的i。考虑图1-29所示的序列。
当j=5时，B5=12，因此目标是找一个Bi-1≤12-7=5的最大i。注意到B是递增的（别忘了，本题中所有Ai均为整数），所以可以用二分查找。
【代码】
/*********************************
* 日期：2014-5-14
* 作者：SJF0115
* 题号: 1121 - Subsequence
* 地址：http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=246&page=show_problem&problem=3562
* 来源：UVA
* 结果：Accepted
**********************************/
#include <iostream>
#include <stdio.h>
#include <algorithm>
#include <string.h>
using namespace std;
#define N 100001
int A[N];
int B[N];
//二分查找最接近target但不大于target
int BinarySerach(int target,int R){
int L = 0;
int mid = 0;
while(L < R){
mid = L + (R - L) / 2;
if(B[mid] > target){
R = mid;
}
else{
L = mid + 1;
}
}
return L;
}
int main(){
int n,s,i,j;
//freopen("C:\\Users\\wt\\Desktop\\acm.txt","r",stdin);
while(scanf("%d %d",&n,&s) != EOF){
int minLen = n+1;
B[0] = 0;
for(i = 1;i <= n;i++){
scanf("%d",&A[i]);
//序列前缀和
B[i] = B[i-1] + A[i];
}
for(j = 1;j <= n;j++){
int target = B[j] - s;
//二分查找
int index = BinarySerach(target,j-1);
if(index > 0){
minLen = min(minLen,j-index+1);
}
}
//没有满足条件的序列
if(minLen == n+1){
minLen = 0;
}
cout<<minLen<<endl;
}//while
return 0;
}
