1104 Sum of Number Segments （20 分)

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1104 Sum of Number Segments （20 分)

Given a sequence of positive numbers, a segment is defined to be a consecutive subsequence. For example, given the sequence { 0.1, 0.2, 0.3, 0.4 }, we have 10 segments: (0.1) (0.1, 0.2) (0.1, 0.2, 0.3) (0.1, 0.2, 0.3, 0.4) (0.2) (0.2, 0.3) (0.2, 0.3, 0.4) (0.3) (0.3, 0.4) and (0.4).

Now given a sequence, you are supposed to find the sum of all the numbers in all the segments. For the previous example, the sum of all the 10 segments is 0.1 + 0.3 + 0.6 + 1.0 + 0.2 + 0.5 + 0.9 + 0.3 + 0.7 + 0.4 = 5.0.

Input Specification:
Each input file contains one test case. For each case, the first line gives a positive integer N, the size of the sequence which is no more than 10⁵. The next line contains N positive numbers in the sequence, each no more than 1.0, separated by a space.

Output Specification:
For each test case, print in one line the sum of all the numbers in all the segments, accurate up to 2 decimal places.

Sample Input:
4
0.1 0.2 0.3 0.4
Sample Output:
5.00

根据题目算法得到的超时代码

#include <iostream>
#include <cstdio>
using namespace std;
int main() {
	int N;
	scanf("%d", &N);
	double *Nums = new double[N];	//根据所给数字大小开辟数组
	for (int i = 0; i < N; i++) scanf("%lf", &Nums[i]);	//将所有数字存入数组
	double Sum = 0;
	for (int i = 0; i < N; i++) {
		double TmpSum = 0;
		int pos = i;
		while (pos != N) {	//当pos未到达末尾时
			TmpSum += Nums[pos++];	//得到子列和
			Sum += TmpSum;	//总和加子列和
		}
	}
	printf("%.2lf\n", Sum);		//精确到小数点后两位
	delete[] Nums;		//释放数组
	return 0;
}					//此代码后两个测试点超时

优化算法AC代码

#include <iostream>
#include <cstdio>
using namespace std;
int main() {
	int N;
	scanf("%d", &N);
	double *Nums = new double[N];	//根据所给数字大小开辟数组
	for (int i = 0; i < N; i++) scanf("%lf", &Nums[i]);	//将所有数字存入数组
	double Sum = 0;
	for (int i = 0; i < N; i++) Sum += Nums[i] * (i + 1) * (N - i);		//优化算法 建议多写几组统计得到规律
	printf("%.2lf\n", Sum);		//精确到小数点后两位
	delete[] Nums;		//释放数组
	return 0;
}