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 0.4Sample Output:
5.00
題目大意
給你一組數據,讓你求任意兩個位置之間的數字和的和。
題目解析
這種題打個表,找規律就好,每個數字加上的次數都是有規律的。
#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;
}