PAT 1104. Sum of Number Segments (20) double要在int型前面！未解之谜

1104. Sum of Number Segments (20)

时间限制

200 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CAO, Peng

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:

40.1 0.2 0.3 0.4 

Sample Output:

5.00

很蛋疼的一道题。坑点在于数据溢出。

#include <iostream>#include<stdio.h>#include<string>#include<queue>#include<algorithm>#include<string.h>#include<vector>#include<map>#include<math.h>using namespace std;double num[100000]={0};int main(){    int n;    cin>>n;    for(int i=1;i<=n;i++)    cin>>num[i];    double sum=num[0];    for(int i=1;i<=n;i++)    {        sum+=num[i]*i*(n+1-i);    //num[]一定要放最前面！！！    }        printf("%.2lf",sum);    //system("pause");  return 0;}

我用动态规划做的只能过三个，不知道为什么

#include <iostream>#include<stdio.h>#include<string>#include<queue>#include<algorithm>#include<string.h>#include<vector>#include<map>#include<math.h>using namespace std;double num[100002]={0};double dp[100002]={0};long long cnt[100002]={0};int main(){    int n;    cin>>n;    for(int i=0;i<n;i++)    cin>>num[i];    dp[0]=num[0];      //以i为结尾的串的和    cnt[0]=1;          //以i为结尾的串的数量    double sum=num[0];    for(int i=1;i<n;i++)    {      dp[i]=num[i]*cnt[i-1]+dp[i-1];        cnt[i]=cnt[i-1]+1;        dp[i]+=num[i];      sum+=dp[i];    }        printf("%.2lf",sum);    //system("pause");  return 0;}