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[AHOI2009]飞行棋 BZOJ1800

Raow1 2022-05-27 阅读 105

题目描述

给出圆周上的若干个点,已知点与点之间的弧长,其值均为正整数,并依圆周顺序排列。 请找出这些点中有没有可以围成矩形的,并希望在最短时间内找出所有不重复矩形。

输入输出格式

输入格式:

第一行为正整数N,表示点的个数,接下来N行分别为这N个点所分割的各个圆弧长度

输出格式:

所构成不重复矩形的个数

输入输出样例

输入样例#1:

复制

8
1
2
2
3
1
1
3
3

输出样例#1: 复制

3

说明

N<=20 [AHOI2009]飞行棋 BZOJ1800_#define

 

#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
//#include
//#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 pii;
#define pi acos(-1.0)
//const int N = 1005;
#define REP(i,n) for(int i=0;i<(n);i++)
typedef pair 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;
}

 

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