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1104 Sum of Number Segments

Statement

Metadata

  • 作者: CAO, Peng
  • 单位: Google
  • 代码长度限制: 16 KB
  • 时间限制: 200 ms
  • 内存限制: 64 MB

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 10^5. 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

Thanks to Ruihan Zheng for correcting the test data.

Solution

#include <ctype.h>
#include <algorithm>
#include <climits>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <ctime>
#include <deque>
#include <iomanip>
#include <iostream>
#include <limits>
#include <map>
#include <numeric>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <string>
#include <vector>

#define CLR(a) memset(a, 0, sizeof(a))
#define pb push_back

using namespace std;
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;

const double PI = 3.14159265358979323846264338327;
const double E = exp(1);
const double eps = 1e-6;

const int INF = 0x3f3f3f3f;
const int maxn = 1e5 + 5;
const int MOD = 1e9 + 7;

int main() {
    int n;
    double num;
    scanf("%d", &n);
    double sum = 0.0;
    for (int i = 0; i < n; i++) {
        scanf("%lf", &num);
        sum += num * (i + 1) * (n - i);
    }
    printf("%.2lf\n", sum);
}
Last update: May 4, 2022
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