A1104. Sum of Number Segments

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) (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

 

#include<cstdio>
#include<iostream>
#include<algorithm>
using namespace std;
int main(){
    int N;
    double ans = 0, temp;
    scanf("%d", &N);
    for(int i = 1; i <= N; i++){
        scanf("%lf", &temp);
        ans += 1.0 * i * (N + 1 - i) * temp;
    }
    printf("%.2lf", ans);
    cin >> N;
    return 0;
}

 

总结：

1、主要是找规律， 计算出每个数一共出现几次（与位置有关），然后做乘法再累加即可。第i个数（从1开始）出现的次数是：i * (n + 1 - i)。

2、i * （N + 1 - i）有可能超过int的范围。 所以做类型转换时，1.0应该乘在式子的最前面。否则当int溢出后再做转换就无效了。