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
一,思路问题
1,我的思路是先将总的片段和加起来乘以元素个数,然后依次相减,规律是第一个元素乘以N减去N-1次,第二个元素乘以N-1减去N-2次,以此类推。但是程序健壮性不强,易出错。
2,更好的方法是推出每一位的元素出现的次数,通过每一个长度的片段这个元素出现的次数容易得到,得出公式,第i位元素出现的次数为i*(N-i+1).
二,正确代码
#include<cstdio>
using namespace std;
const int max_n = 100100;
double arr[max_n];
int main() {
int N = 0;
double v = 0, ans = 0;
scanf("%d", &N);
for (int i = 1; i <= N; i++) {
scanf("%lf", &v);
ans += v*i*(N - i + 1);
}
printf("%.2f", ans);
return 0;
}