Subsequence

Time Limit: 1000MS	Memory Limit: 65536K
Total Submissions: 10613	Accepted: 4396

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

題目大意：

       給出N個數字，每個數字不大於10000，給出一個S，在N個數字中挑選出連續的a的數字，使這a個數字的和大於或等於S。請問這個a最小是幾？

遇到的問題和解題思路：

       這道題目是挑戰程式設計146頁的例題。書上給出了兩種解法，不過我稍微改了一下下，但是思路還是一樣的。

給出兩個程式碼：

//二分法 #include<cstdio> #include<algorithm> #include<cstring> using namespace std; int a[100000 + 10]; int n, s; void work(){ int sum = 0; for (int i = 0; i < n; i++){ sum += a[i]; a[i] = sum; } //for (int i = 0; i < n; i++)printf(" %d ", a[i]); int res = n + 1; for (int i = 0; a[i] + s <= a[n - 1]; i++){ int tmp = lower_bound(a + i, a + n, a[i] + s) - a - i; //printf("tmp = %d\n", tmp); res = min(res, tmp); } if (res == n + 1)printf("0\n"); else printf("%d\n", res); } int main(){ int t; scanf("%d", &t); while (t--){ memset(a, 0, sizeof(a)); scanf("%d%d", &n, &s); for (int i = 0; i < n; i++)scanf("%d", a + i); work(); } return 0; } //尺取法 #include<cstdio> #include<algorithm> #include<cstring> using namespace std; int n, a[100000 + 10], s; void solve(){ /*for (int i = 0; i < n; i++){ printf("%d ", a[i]); } printf("\n");*/ int left = 0, right = 0, sum = 0, res = n + 1; while (true){ while (right < n && sum < s){ sum += a[right++]; } if (sum < s) break; //printf("mid = %d\n", right - left); res = min(right - left, res); sum -= a[left++]; } if (res < n + 1)printf("%d\n", res); else printf("0\n"); } int main(){ int t; scanf("%d", &t); while(t--){ memset(a, 0, sizeof(a)); scanf("%d%d", &n, &s); for (int i = 0; i < n; i++){ scanf("%d", a + i); } solve(); } return 0; } </cstring></algorithm></cstdio></cstring></algorithm></cstdio>