2017 UESTC Training for Math

2017 UESTC Training for Math

A    sg博弈水题

#include<bits/stdc++.h>
using namespace std;
#pragma comment(linker, "/STACK:102400000,102400000")
#define rep(i,a,b) for (int i=a; i<=b; ++i)
#define per(i,b,a) for (int i=b; i>=a; --i)
#define mes(a,b)  memset(a,b,sizeof(a))
#define INF 0x3f3f3f3f
#define MP make_pair
#define PB push_back
#define fi  first
#define se  second
typedef long long ll;
const int N = 10005;

int k, s[N], m, mi, a[N], sg[N];
bool vis[N];
void getSG()
{
    mes(sg, 0);
    rep(i,0,N-1)
    {
        mes(vis, 0);
        rep(j,1,k) if(i-s[j]>=0)
            vis[sg[i-s[j]]]=1;
        rep(j,0,N-1) if(vis[j]==0) {
            sg[i]=j;
            break;
        }
    }
}
int main()
{
    scanf("%d", &k);
    rep(i,1,k) scanf("%d", &s[i]);
    getSG();
    scanf("%d", &m);
    rep(i,1,m)
    {
        scanf("%d", &mi);
        int ans=0;
        rep(j,1,mi) {
            scanf("%d", &a[i]);
            ans ^= sg[a[i]];
        }
        if(ans==0) puts("lose!");
        else puts("win!");
    }

    return 0;
}

B    求两圆相交的面积模板

#define  PI  acos(-1.0)
struct Circle { double x, y, r; };
double dis(Circle a, Circle b) {
    return  sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}
double IntersectionArea_TwoCircles(Circle c1, Circle c2)
{
    double s = dis(c1,c2);
    if(c1.r<c2.r) swap(c1, c2);
    if(c1.r+c2.r <= s) return 0;
    else if(s <= c1.r-c2.r) return PI*c2.r*c2.r;
    else {
        double ang1 = acos((c1.r*c1.r+s*s-c2.r*c2.r)/(2*c1.r*s));
        double ang2 = acos((c2.r*c2.r+s*s-c1.r*c1.r)/(2*c2.r*s));
        return ang1*c1.r*c1.r + ang2*c2.r*c2.r - c2.r*s*sin(ang2);
    }
}

 

E    水题

#include<bits/stdc++.h>
using namespace std;
#pragma comment(linker, "/STACK:102400000,102400000")
#define rep(i,a,b) for (int i=a; i<=b; ++i)
#define per(i,b,a) for (int i=b; i>=a; --i)
#define mes(a,b)  memset(a,b,sizeof(a))
#define INF 0x3f3f3f3f
#define MP make_pair
#define PB push_back
#define fi  first
#define se  second
typedef long long ll;
const int N = 200005;

ll  A(int n, int m)
{
    ll  ans = 1;
    rep(i,n-m+1,n) ans *= i;
    return ans;
}
int main()
{
    int n;
    scanf("%d", &n);
    printf("%lld\n", A(n,n)/n*A(n,n));

    return 0;
}

L    第二类斯特林数

题意： n 个人放在 k个相同的篝火中，问有多少种方案。

tags：参考大神博客   

类似于dp递推，dp[i][j]表示 i 个物体放入 j 个盒子的方案数，则 dp[i][j] = j * dp[i-1][j] + dp[i-1][j-1] 。

#include<bits/stdc++.h>
using namespace std;
#pragma comment(linker, "/STACK:102400000,102400000")
#define rep(i,a,b) for (int i=a; i<=b; ++i)
#define per(i,b,a) for (int i=b; i>=a; --i)
#define mes(a,b)  memset(a,b,sizeof(a))
#define INF 0x3f3f3f3f
#define MP make_pair
#define PB push_back
#define fi  first
#define se  second
typedef long long ll;
const int N = 1005, mod = 1e9+7;

ll dp[N][N], n, k;
int main()
{
    rep(i,1,N-1) dp[i][1]=1;
    rep(i,2,N-1) rep(j,1,N-1)
    {
        dp[i][j] = (j*dp[i-1][j]+dp[i-1][j-1]) %mod;
    }
    int T;  scanf("%d", &T);
    while(T--)
    {
        scanf("%lld %lld", &n, &k);
        printf("%lld\n", (dp[n][k]+mod)%mod);
    }

    return 0;
}