5.1 PAT A1104 Sum of Number Segments (20分)(数列子序列和——找规律）

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 10​5​​. 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

此题直接暴力会超时，所以要找出规律。每一个位置出现的次数为i*（n+1-i)次。找出这个规律后，此题便迎刃而解了。

直接暴力超时代码：

//此题直接暴力枚举会超时 
#include <cstdio> 
double n[100010];
int main()
{
	int m;
	while(scanf("%d",&m)!=EOF)
	{
		for(int i=0;i<m;++i)
			scanf("%lf",&n[i]);
		double ans=0.0;
		int j,i=0;
		while(i<m)
		{
			j=i;
			while(j<m)
			{
				for(int k=i;k<=j;++k)
					ans+=n[k];
				j++;
			}
			i++; 
		}
		printf("%.2f\n",ans);
	}
	return 0;
}

 AC代码：

#include <cstdio> 
//找每位出现的次数的规律 
int main()
{
	int n;
	while(scanf("%d",&n)!=EOF)
	{
		double v,ans=0;
		for(int i=1;i<=n;++i)
		{
			scanf("%lf",&v);
			ans+=v*i*(n+1-i); //第i位出现的次数为 i*(n+1-i)次 
		} 
		printf("%.2f\n",ans);
	}
	return 0;
}