1104 Sum of Number Segments (20 分)有超时
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
最后两个样例会超时
#include<iostream> #include<vector> #include<algorithm> #include<map> #include<set> #include<cmath> #include<climits> #include<sstream> #include<cstdio> #include<string.h> #include<unordered_map> using namespace std; long long int maxn=1000000; int main() { int n; scanf("%d",&n); double a; double sum[n+1]; memset(sum,0,sizeof(sum)); double result=0; for(int i=1;i<n+1;i++) { scanf("%lf",&a); sum[i]=sum[i-1]+a; result+=sum[i]; } for(int i=1;i<n;i++) { for(int j=i+1;j<n+1;j++) { result+=sum[j]-sum[i]; //cout<<sum[j]-sum[i]<<endl; } } printf("%.2f",result); return 0; }
posted on 2019-02-01 19:20 ZhangのBlog 阅读(142) 评论(0) 编辑 收藏 举报