尺取 Poj3601
http://poj.org/problem?id=3061
Subsequence
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 20744 | Accepted: 8849 |
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 5Sample Output
2
3题意：给定长度为n的数列整数a0,a1,a2,a3 ..... an-1以及整数S。求出综合不小于S的连续子序列的长度的最小值。如果解不存在，则输出0。
思路：其实相当于当不满足条件就入队，然后得到队列长度，再将队首元素出队，再进行下一次的入队，直到满足条件再次出队，并且将这一次的长度与历史最短长度进行取舍，最后扫到最后的元素却无法再满足入队条件的时候就结束，此时用O(n)的时间就可以得到答案。
我的ac代码（较混乱）：
#include 模板（用了首尾值）：
#include
#include
#include
#include
#include
#include
typedef long long ll;
using namespace std;
const int INF=0x3f3f3f3f;
int a[100010];
int main(){
ios::sync_with_stdio(false);
int N;
cin>>N;
while(N--){
int n,m;
int ans=INF;
cin>>n>>m;
for(int i=1;i<=n;i++) cin>>a[i];
int i=1,j=1;//i为序列的首，j为序列的尾
int sum = 0;
while(1)
{
while(j<=n&&sum