題目連結

題目描述

給定一個數組a[N],N小於1e5。把陣列劃分成若干個片段，每個片段的和都不為0，問有多少種劃分方法？

方法描述

定義f(i)表示0~i共有多少種劃分方式，則$f(j)=\sum_{i\in[0,j) and sum(a[i+1:j])!=0} f(i)$
相當於統計$f(j)=\sum_{i \in [0,j)} f(i)-\sum_{i \in [0,j) and sum(a[i+1:j])==0} f(i)$。對於此式第二項可以使用map記錄下來。

#include<iostream>
#include<stdio.h>
#include<map> 
using namespace std; 
typedef long long ll;
const int maxn = 1e9 + 7;
const int maxcount = 1e5 + 3;
const int maxvalue = 103; 
int n;
int a[maxcount];
int pre[maxcount];
map<int, int>ma; 
int main() {
    freopen("in.txt", "r", stdin);
    cin >> n;
    for (int i = 0; i < n; i++)scanf("%d", a + i+1); 
    pre[0]=a[0] = 0;
    ma[0] = 1;
    for (int i = 1; i <= n; i++)pre[i] = pre[i - 1] + a[i];  
    ll s = 1;
    ll now = 0;
    for (int i=1; i <= n; i++) {  
        now = (s- ma[pre[i]]+maxn)%maxn;  
        s = (s + now) % maxn; 
        if (ma.count(pre[i]) == 0)ma[pre[i]] = 0;
        ma[pre[i]] = (ma[pre[i]]+now)%maxn;
    } 
    cout << now<< endl;
    return 0;
}