[AHOI2009]飛行棋 BZOJ1800

題目描述

給出圓周上的若干個點，已知點與點之間的弧長，其值均為正整數，并依圓周順序排列。 請找出這些點中有沒有可以圍成矩形的，并希望在最短時間內找出所有不重復矩形。

輸入輸出格式

輸入格式：

第一行為正整數N，表示點的個數，接下來N行分別為這N個點所分割的各個圓弧長度

輸出格式：

所構成不重復矩形的個數

輸入輸出樣例

輸入樣例#1： 復制

8
1
2
2
3
1
1
3
3

輸出樣例#1： 復制

3

說明

N<=20

?

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<bitset>
#include<ctime>
#include<deque>
#include<stack>
#include<functional>
#include<sstream>
//#include<cctype>
//#pragma GCC optimize(2)
using namespace std;
#define maxn 200005
#define inf 0x7fffffff
//#define INF 1e18
#define rdint(x) scanf("%d",&x)
#define rdllt(x) scanf("%lld",&x)
#define rdult(x) scanf("%lu",&x)
#define rdlf(x) scanf("%lf",&x)
#define rdstr(x) scanf("%s",x)
typedef long long  ll;
typedef unsigned long long ull;
typedef unsigned int U;
#define ms(x) memset((x),0,sizeof(x))
const long long int mod = 1e9 + 7;
#define Mod 1000000000
#define sq(x) (x)*(x)
#define eps 1e-3
typedef pair<int, int> pii;
#define pi acos(-1.0)
//const int N = 1005;
#define REP(i,n) for(int i=0;i<(n);i++)
typedef pair<int, int> pii;
inline ll rd() {ll x = 0;char c = getchar();bool f = false;while (!isdigit(c)) {if (c == '-') f = true;c = getchar();}while (isdigit(c)) {x = (x << 1) + (x << 3) + (c ^ 48);c = getchar();}return f ? -x : x;
}ll gcd(ll a, ll b) {return b == 0 ? a : gcd(b, a%b);
}
int sqr(int x) { return x * x; }/*ll ans;
ll exgcd(ll a, ll b, ll &x, ll &y) {if (!b) {x = 1; y = 0; return a;}ans = exgcd(b, a%b, x, y);ll t = x; x = y; y = t - a / b * y;return ans;
}
*/int n;
int a[maxn];
int sum[maxn];int main() {//ios::sync_with_stdio(0);rdint(n);for (int i = 1; i <= n; i++)rdint(a[i]), sum[i] = sum[i - 1] + a[i];int ans = 0;for (int i = 1; i <= n; i++) {for (int j = i + 1; j <= n; j++) {if (sum[j] - sum[i] == (sum[n] / 2))ans++;}}cout << ans * (ans - 1) / 2 << endl;return 0;
}

?