PAT Sum of Number Segments[数学问题][一般]

1104 Sum of Number Segments（20 分）

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

 题目大意：给出一个序列，找出所有的子序列，并对子序列求和输出。

 //首先就想到用树状数组，所以写了一个树状数组巩固一下，其中遇到了一些小问题，解决之后提交pat发现运行超时，只通过了两个测试点，应该是不能用这个来做了。

#include <iostream>
#include <stdio.h>
using namespace std;
float a[100001];
int lowbit(int x){
    return x&-x;
}

void update(int index,float x){//为什么你会死循环呢？
    //因为！！树状数组下标应该从1开始的！！！
    for(int i=index;i<=100001;i+=lowbit(i)){
        a[i]+=x;
    }
}

float getsum(int x){
    float ret=0.0;
    for(int i=x;i>0;i-=lowbit(i)){
        ret+=a[i];
    }
    return ret;
}

int main() {
    int n;
    scanf("%d",&n);
    float x;
    for(int i=1;i<=n;i++){
        scanf("%f",&x);
        update(i,x);
    }
//    for(int i=1;i<=4;i++)
//        printf("%f ",a[i]);
    float sum=0.0;
    for(int i=1;i<=n;i++){
        for(int j=i;j<=n;j++){
            if(i==1){
                sum+=getsum(j);
            }else {
                sum+=(getsum(j)-getsum(i-1));
            }

        }
    }
    printf("%.2f",sum);
    return 0;
}

//基本上是O（n^2）的复杂度。

小问题：1.getsum函数没有对ret进行返回。。。导致sum一直输出nan....

2.树状数组下标应该从1开始！不是0，0会导致update死循环。

代码来自：https://www.liuchuo.net/archives/1921

#include <iostream>
using namespace std;
int main() {
    int n;
    cin >> n;
    double a[100001];
    double sum = 0.0;
    for (int i = 1; i <= n; i++) {
        cin >> a[i];
        sum = sum + a[i] * i * (n - i + 1);//这里计算每个数出现的次数！。
        //是一个找规律的问题。
    }
    printf("%.2f", sum);
    return 0;
}

//当我看到这个题的代码如此短小精悍，惊呆了。

//还是有点不明白，搜索了一下：

对于第i个数字(i=0~n-1)，它每组出现的次数为n-i，出现在前i+1个组中

 非常厉害！学习了！