PAT 甲级 1104 Sum of Number Segments (20分)(有坑,int *int 可能会溢出)
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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
题意:
给定一个正数数列,我们可以从中截取任意的连续的几个数,称为片段。例如,给定数列{0.1, 0.2, 0.3, 0.4},我们有(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) 这10个片段。给定正整数数列,求出全部片段包含的所有的数之和。如本例中10个片段总和是0.1 + 0.3 + 0.6 + 1.0 + 0.2 + 0.5 + 0.9 + 0.3 + 0.7 + 0.4 = 5.0,在一行中输出该序列所有片段包含的数之和,精确到小数点后2位~
题解:
将数列中的每个数字读取到temp中,假设我们选取的片段中包括temp,且这个片段的首尾指针分别为p和q,那么对于p,有i种选择,即12…i,对于q,有n-i+1种选择,即i, i+1, … n,所以p和q组合形成的首尾片段有i * (n-i+1)种,因为每个里面都会出现temp,所以temp引起的总和为temp * i * (n – i + 1);遍历完所有数字,将每个temp引起的总和都累加到sum中,最后输出sum的值~
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版权声明:本文为CSDN博主「柳婼」的原创文章,遵循 CC 4.0 BY-SA 版权协议,转载请附上原文出处链接及本声明。
原文链接:https://blog.csdn.net/liuchuo/article/details/51985824
真的很细节 挖了好多次。 int*int*double 转换是 int*int 变为 int ,然后 int *double
但是 当 i 去100000时 int *int 可能会溢出,然后就答案错误,解决方法要么把int换为long long 要么把 double 提前
使得 double * int 变为 double
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版权声明:本文为CSDN博主「Cute_jinx」的原创文章,遵循 CC 4.0 BY-SA 版权协议,转载请附上原文出处链接及本声明。
原文链接:https://blog.csdn.net/Cute_jinx/article/details/82497800
AC代码:
#include<bits/stdc++.h> using namespace std; double a; int n; double s=0.0; int main(){ cin>>n; for(int i=1;i<=n;i++){ cin>>a; s+=a*i*(n-i+1); //int*int*double 转换是 int*int 变为 int ,然后 int *double //当 i 去100000时 int *int 可能会溢出,然后就答案错误, //解决方法要么把int换为long long 要么把 double 提前 使得 double * int 变为 double //s+=(n-i+1)*i*a最后两个测试点过不了 } printf("%.2f",s); return 0; }
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