zoukankan      html  css  js  c++  java
  • 1104 Sum of Number Segments (20分)(long double)

    1104 Sum of Number Segments (20分)

     

    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

    long double的输出方式为 "%Lf"

     1 #include <cstdio>
     2 using namespace std;
     3 
     4 const int maxn = 1e5 + 5;
     5 int n;
     6 double val[maxn];
     7 
     8 int main() {
     9     scanf("%d", &n);
    10     for(int i = 0; i < n; i ++) {
    11         scanf("%lf", &val[i]);
    12     }
    13     long double sum = 0;
    14     for(int i = 0; i < n; i ++) {
    15         sum += val[i] * (i + 1) * (n - i);
    16     }
    17     printf("%.2Lf
    ", sum);
    18     return 0;
    19 }
     
  • 相关阅读:
    HDU4366 Successor 线段树+预处理
    POJ2823 Sliding Window 单调队列
    HDU寻找最大值 递推求连续区间
    UVA846 Steps 二分查找
    HDU3415 Max Sum of MaxKsubsequence 单调队列
    HDU时间挑战 树状数组
    UVA10168 Summation of Four Primes 哥德巴赫猜想
    UESTC我要长高 DP优化
    HDUChess 递推
    HDU4362 Dragon Ball DP+优化
  • 原文地址:https://www.cnblogs.com/bianjunting/p/13211081.html
Copyright © 2011-2022 走看看