LA 2678 Subsequence .

题目地址：http://vjudge.net/problem/UVALive-2678

类似于一种滑动窗口

i~j始终维护<=S的最短子序列

先j向右扫描，i不动，直到sum=∑a[k] (i<=k<=j) >=S;

然后i向右扫描，直到刚好sum>=S，更新答案ans=min(ans,j-i)

算法复杂度O(n)

#include <bits/stdc++.h>
using namespace std;
#define REP(i,a,b) for(int i=a;i<=(b);++i)
#define REPD(i,a,b) for(int i=a;i>=(b);--i)
#define max(a,b) ((a)>(b)?(a):(b))
#define min(a,b) ((a)<(b)?(a):(b))
const int maxn=100000+5,INF=0x3f3f3f3f;
int a[maxn];
int main(int argc, char const *argv[])
{
int N,S;
while(scanf("%d%d",&N,&S)==2){
REP(i,1,N) scanf("%d",&a[i]);
int i=0,ans=INF,sum=0;
REP(j,1,N){
sum+=a[j];
if(sum<=S) continue;
while(sum-a[i+1]>=S) sum-=a[++i]; //sum+a[j]>=S;
ans=min(ans,j-i);
}
printf("%d\n", ans==INF?0:ans);
}
return 0;
}