PAT 1104 Sum of Number Segments
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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 1. 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<bits/stdc++.h> using namespace std; typedef long long ll; int main(){ int n;cin >> n; double sum = 0.0; for(int i=1;i <= n;i++){ double x;cin >> x; sum+=i*(n+1-i)*x; } printf("%.2f",sum); return 0; }
x放右面就过不了几个点,x放左边就能过了
#include<bits/stdc++.h> using namespace std; typedef long long ll; int main(){ int n;cin >> n; double sum = 0.0; for(int i=1;i <= n;i++){ double x;cin >> x; sum+=x*i*(n+1-i); } printf("%.2f",sum); return 0; }
对拍了半小时也没啥不同,这个bug先欠着,以后记着把double放左边先乘
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