A1104 Sum of Number Segments (20 point(s))

简单推理

1. 原文

原题

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

2. 解析

• 1x4x0.1

• 2x3x0.2

• 3x2x0.3

• 4x1x0.4

• (i+1)x(n-i)xnum[i]

  测试点2 double会超时

3. AC代码

#include<cstdio>
double num[100010];
int main(){
	int n;
	scanf("%d",&n);
	
	long long ans=0;
	double a;
	for (int i = 0; i < n; ++i)
	{
		scanf("%lf",&a);
		ans+=(long long)(a*1000)*(i+1)*(n-i);
	}
	printf("%.2f\n", ans/1000.0);
	return 0;
}

Way2:

#include<cstdio>
long double num[100010];
int main(){
	freopen("A1104.txt","r",stdin);
	int n;
	scanf("%d",&n);
	for (int i = 0; i < n; ++i)
	{
		scanf("%Lf",&num[i]);
	}
	long double ans = 0;
	for (int i = 0; i < n; ++i)
	{
		ans+=num[i]*(i+1)*(n-i);
	}
	printf("%.2Lf\n",ans);
	return 0;
}