国产探花免费观看_亚洲丰满少妇自慰呻吟_97日韩有码在线_资源在线日韩欧美_一区二区精品毛片,辰东完美世界有声小说,欢乐颂第一季,yy玄幻小说排行榜完本

首頁 > 學院 > 開發設計 > 正文

PAT A1104. Sum of Number Segments (20)

2019-11-08 02:04:31
字體:
來源:轉載
供稿:網友

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
#include <cstdio>#include <algorithm>#include <cstring>#include <string>#define Max 100010using namespace std;int main(){	double m=0,x;	int n;	scanf("%d",&n);	for(int i = 1 ; i <= n ; i ++)	{		scanf("%lf",&x);	    m+=x*i*(n-i+1);	}	printf("%.2f/n",m);	system("pause");	return 0;}
發表評論 共有條評論
用戶名: 密碼:
驗證碼: 匿名發表
主站蜘蛛池模板: 平和县| 安福县| 开平市| 阿荣旗| 长武县| 宜章县| 弥勒县| 汤阴县| 贺州市| 开远市| 建始县| 三穗县| 咸丰县| 白城市| 宝山区| 扎赉特旗| 巴中市| 醴陵市| 临颍县| 东乌珠穆沁旗| 大名县| 淮北市| 马公市| 射阳县| 大关县| 沂南县| 揭阳市| 丰镇市| 栾城县| 克东县| 龙游县| 凌海市| 漾濞| 凤凰县| 武宁县| 元氏县| 南丰县| 水城县| 剑川县| 武城县| 太仓市|