題目大意就是問能分出幾個使三個區間的和相等的分發

 1 #include <algorithm>
 2 #include <stack>
 3 #include <istream>
 4 #include <stdio.h>
 5 #include <map>
 6 #include <math.h>
 7 #include <vector>
 8 #include <iostream>
 9 #include <queue>
10 #include <string.h>
11
 
 #include <set>
12 #include <cstdio>
13 #define FR(i,n) for(int i=0;i<n;i++)
14 #define MAX 2005
15 #define mkp pair <int,int>
16 using namespace std;
17 #include <bits/stdc++.h>
18 const int maxn = 5e5 + 40;
19 typedef long long ll;
20 const int  inf = 0x3fffff;
21 void read(ll &x) {
 
22     char ch; bool flag = 0;
23     for (ch = getchar(); !isdigit(ch) && ((flag |= (ch == ‘-‘)) || 1); ch = getchar());
24     for (x = 0; isdigit(ch); x = (x << 1) + (x << 3) + ch - 48, ch = getchar());
25     x *= 1 - 2 * flag;
26 }
27 
28 
29 ll n;
30 ll sum[maxn],arr[maxn];
31
 
 ll val[maxn];
32 ll s2[maxn];
33 ll l[maxn];
34 int main() {
35     read(n);
36     for(int i=0;i<n;i++)read(val[i]);
37     sum[0]=val[0];
38     for(int i=1;i<n;i++)sum[i]+=sum[i-1]+val[i];
39     //int i = 0,j = n-1;
40     for(int i=n-1;i>=0;i--)s2[i]+=s2[i+1]+val[i];
41     if(sum[n-1]%3)return 0*printf("%d\n",0);
42     ll temp = 0;
43     for(int i=0;i<n;i++){ l[i]=temp;
44         if(sum[i]==sum[n-1]/3)temp++;
45 
46     }
47     ll ans = 0;
48     for(int i=n-1;i>=1;i--){
49         if(s2[i]==sum[n-1]/3){
50             ans+=l[i-1];
51         }
52        // cout<<ans<<endl;
53     }
54     cout<<ans<<endl;
55     return 0;
56 }

C. Number of Ways 思維