The land is for sale in CyberCity, and is divided into several pieces. Here it is assumed that each piece of land has exactly two neighboring pieces, except the first and the last that have only one. One can buy several contiguous（连续的） pieces at a time. Now given the list of prices of the land pieces, your job is to tell a customer in how many different ways that he/she can buy with a certain amount of money.
Input Specification:
Each input file contains one test case. Each case first gives in a line two positive integers: N (≤10^4), the number of pieces of the land (hence the land pieces are numbered from 1 to N in order), and M (≤10^9), the amount of money that your customer has.
Then in the next line, N positive integers are given, where the i-th one is the price of the i-th piece of the land.
It is guaranteed that the total price of the land is no more than 10^9.
Output Specification:
For each test case, print the number of different ways that your customer can buy. Notice that the pieces must be contiguous.
Sample Input:
5 85
38 42 15 24 9
Sample Output:
11
Hint:
The 11 different ways are:
38
42
15
24
9
38 42
42 15
42 15 24
15 24
15 24 9
24 9
题目大意：数码城市有土地出售。待售的土地被划分成若干块，每一块标有一个价格。这里假设每块土地只有两块相邻的土地，除了开头和结尾的两块是只有一块邻居的。每位客户可以购买多块连续相邻的土地。现给定这一系列土地的标价，请你编写程序，根据客户手头的现金量，告诉客户有多少种不同的购买方案。
分析：使用pre数组保存土地价格前缀和，pre[i]表示从第一个土地到第i个土地的总价。在计算从第j个房子到第i个土地的总价的时候，只需要计算pre[i]-pre[j-1]即可。两个for循环，枚举所有i与j的值，然后计算出可以购买的方案数量即可。喵喵喵~
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#include <iostream> using namespace std; int N, M, ans, a, pre[10005]; int main(){ cin >> N >> M; for (int i = 1; i <= N; i++) { cin >> a; pre[i] = pre[i - 1] + a; } for (int i = 1; i <= N; i++) { for (int j = i; j <= N; j++) { if (pre[j] - pre[i - 1] <= M) ++ans; else break; } } cout << ans; return 0; } |
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