PAT甲级 1104. Sum of Number Segments (20)

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本文介绍了一个算法问题，即计算给定正数序列中所有连续子序列（片段）内数字的总和。通过分析规律，文章提供了一种高效计算的方法，并附带了C++实现代码。

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

题目大意

给你一组数据，让你求任意两个位置之间的数字和的和。

题目解析

这种题打个表，找规律就好，每个数字加上的次数都是有规律的。

#include <cstdio>
#include <cmath>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <cstdlib>
#include <climits>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

#define ll long long
const int MAXN = 100000 + 5;
const int MAXM = 100000 + 5;
const int INF = 0x7f7f7f7f;
const int dir[][2] = {0,1,1,0,0,-1,-1,0};
template <class XSD> inline XSD f_min(XSD a, XSD b) { if (a > b) a = b; return a; }
template <class XSD> inline XSD f_max(XSD a, XSD b) { if (a < b) a = b; return a; }
bool cmp(int x, int y){return x>y;}
int n, m;

void Getdata(){
}
void Solve(){
    double a[MAXN], sum = 0.0;
    for(int i=0; i<n; i++) scanf("%lf", &a[i]);
    for(int i=0; i<n; i++) sum += a[i]*(n-i)*(i+1);
    printf("%.2f\n",sum);
}
int main(){
    while(~scanf("%d", &n)){
        Getdata();
        Solve();
    }
    return 0;
}