1104. Sum of Number Segments (20)

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:

40.1 0.2 0.3 0.4 

Sample Output:

5.00

这题就是数学问题。可以求出来每个数出现的次数

例如样例

0.1从前面可以出现0次，后面出现4次

0.2从前面出现3次，后面出现3次

0.3从前面出现2+2次（0.1和0.2都包含），后面出现2次

0.4从前面出现1+1+1次，后面出现1次

这样

0<-0.1->4   i=1

3<-0.2->3   i=2

2+2<-0.3->2   i=3

1+1+1<-0.4->1   i=4

就能得到出现的次数为((n-i+1)*(i-1))+n-i+1

我当时也没在意，能合并。。。。就提交这个答案，错两个点

百度搜答案发现提取公因式n-i-1就是i*(n-i-1)

然后在提交就对了。。。。

#include<cstdio>#include<cmath>#include<algorithm>#include<iostream>#include<cstring>#include<queue>#include<vector>#include<set>#include<map>#include<stack>using namespace std;int main(){int n;cin>>n;double a[100010]={0};double sum=0;for(int i=1;i<=n;i++) {cin>>a[i];sum+=a[i]*i*(n-i+1);}printf("%.2f",sum);return 0;}