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 这题不小心溢出了... #include<stdio.h> int main(){ int n,i; scanf("%d",&n); double a[n]; for(i=0;i<n;i++){ scanf("%lf",&a[i]); } double sum=0; for(i=1;i<=n;i++){ // sum=sum+a[i-1]*((n-i+1)*(i));//测试点 2 3 过不了... sum=sum+a[i-1]*(n-i+1)*(i);//全对... // if((n-i+1)*(i)<0){ 卧槽... (n-i+1)*(i)还会溢出... 坑比... // while(1); // } } printf("%.2lf\n",sum); }