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.
解题思路
首先想到的自然是遍历咯,双指针,复杂度O( n^2 ),然后提交有两个样例超时;然后是数学规律解题,现分析如下,假设现在只有4个数,分别是1,2,3,4.
(1)(1,2)(1,2,3),(1,2,3,4) ----以1开头的片段
(2)(2,3)(2,3,4) ----以2开头的片段
(3)(3,4) ----以3开头的片段
(4) ----以4开头的片段
现在考虑1出现了多少次,1只可能出现在以1开头的片段中,不难得出其出现次数为数组的长度4;2除了可能出现在以2开头的片段中(3次),还可能出现在以1出现的片段中但不包含(1)(3次);3除了可能出现在以3开头的片段中,还可能出现在以1开头的片段中但不包含(1)(1,2)(2次),还可能出现在以2开头的片段中但不包含(2)(2次)。。现在是不是找到规律了?某个数出现在以某数开头的片段中的次数是一样的!接下来就不说了,应该很明了了。
#include <cstdio>
int main(){
/*超时版本
int n;
double list[100010], sum[100010] = {0}, ans = 0;
scanf("%d", &n);
for(int i = 0; i < n; i++){
scanf("%lf", &list[i]);
sum[i + 1] = sum[i] + list[i];
}
int left = 0 , right = 1;
while(left < n){
while(right < n + 1){
ans += sum[right] - sum[left];
right++;
}
left++;
right = left + 1;
}
printf("%.2lf\n", ans);
*/
//规律版本
int n;
double list[100010], sum[100010] = {0}, ans = 0;
scanf("%d", &n);
for(int i = 0; i < n; i++){
scanf("%lf", &list[i]);
ans += list[i] * (i + 1) * (n - i);
}
printf("%.2lf\n", ans);
return 0;
}