2016ACM/ICPC亚洲区大连站-重现赛（感谢大连海事大学）（7/10）

1001题意：n个人，给m对敌对关系，X个好人，Y个坏人。现在问你是否每个人都是要么是好人，要么是坏人。

先看看与X，Y个人有联通的人是否有矛盾，没有矛盾的话咋就继续遍历那些不确定的人关系，随便取一个数3，与其相连的就是4，间隔就要相同，dfs搜过去就可以判断了

#include<bits/stdc++.h>
using namespace std;
#pragma comment(linker, "/STACK:102400000,102400000")
#define ls i<<1
#define rs ls | 1
#define mid ((ll+rr)>>1)
#define pii pair<int,int>
#define MP make_pair
typedef long long LL;
const long long INF = 1e18+1LL;
const double Pi = acos(-1.0);
const int N = 5e5+10, maxn = 1e3+20, mod = 1e9+7, inf = 2e9;

int vis[N],t,head[N],n,m,x,y,z,cal[N],xx[N],yy[N];
int flag;
struct ss{
int to,next;}e[N * 2];
void add(int u,int v) {e[t].next = head[u]; e[t].to = v; head[u] = t++;}

void dfs(int u,int fa) {
        cal[u] = 1;
        for(int i = head[u]; i!=-1; i = e[i].next) {
            int to = e[i].to;
            if(to == fa) continue;
            if(vis[u] == 0) vis[u] = 3;
            if(vis[to] == vis[u]) {
                flag = 1;
                return ;
            }
            if(cal[to]) continue;
            if(vis[u] == 3) vis[to] = 4;
            else if(vis[u] == 4) vis[to] = 3;
            else if(vis[u] == 1) vis[to] = 2;
            else if(vis[u] == 2) vis[to] = 1;
            dfs(to,u);
        }
}

int main() {
        while(scanf("%d%d%d%d",&n,&m,&x,&y)!=EOF) {
                t = 1;
        memset(head,-1,sizeof(head));
            for(int i = 1; i <= m; ++i) {
                int a,b;
                scanf("%d%d",&a,&b);
                add(a,b);
                add(b,a);
            }
            memset(cal,0,sizeof(cal));
            memset(vis,0,sizeof(vis));
            for(int i = 1; i <= x; ++i) scanf("%d",&xx[i]),vis[xx[i]] = 1;
            for(int i = 1; i <= y; ++i) scanf("%d",&yy[i]),vis[yy[i]] = 2;
            flag = 0;
            for(int i = 1; i <= x; ++i) {
                if(!cal[i]) dfs(xx[i],-1);
            }
            for(int i = 1; i <= y; ++i) {
                if(!cal[i]) dfs(yy[i],-1);
            }
            if(flag) {
                puts("NO");
                continue;
            }
            for(int i = 1; i<= n; ++i) {
                if(!cal[i]) {
                    dfs(i,-1);
                }
            }
            for(int i = 1; i <= n; ++i) {
                if(!vis[i]) flag = 1;
            }
            if(flag) puts("NO");
            else puts("YES");
        }
        return 0;
}

1003题意：两堆石子，你可以任选一堆去掉任意个数，你也可从两堆中同时去掉任意个数，最后全部取完的人胜

石子的数量是10^100.高精度的威佐夫博弈，需要黄金比例精确到100位，队友javaA。

import java.util.*;
import java.math.*;

public class Main {
    public static void main(String[] args) {
        Scanner cin = new Scanner(System.in);
        BigInteger k=new BigInteger("6180339887498948482045868343656381177203091798057628621354486227052604628189024497072072041893911374");
        BigInteger p=new BigInteger("10000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000");
        while(cin.hasNext())
        {
            BigInteger nn=cin.nextBigInteger();
            BigInteger mm=cin.nextBigInteger();
            BigInteger n=nn.min(mm);
            BigInteger m=nn.max(mm);
            BigInteger j=n.multiply(k);
            j=j.divide(p);
            BigInteger l=j.multiply(k.add(new BigInteger("1")));
            l=l.divide(p);
            if(n.equals(l)==false)
                j=j.add(new BigInteger("1"));
            n=n.add(j);
            if(n.equals(m))
                System.out.println("0");
            else 
                System.out.println("1");
        }
    }
}

1004题意：给定a,b; 求出满足 LCM(X,Y) = b && X+Y = a的一组解，或者是无解

公式转化：b*gcd(X,Y) = X*Y,X+Y=a;

我们可以知道gcd(X,Y) 必然是a的因子！，那么我们为了枚举gcd(X,Y)就直接去枚举a的因子就是了

枚举以后就相当于求解一个二元一次方程的整数解了；

#include<bits/stdc++.h>
using namespace std;
#pragma comment(linker, "/STACK:102400000,102400000")
#define ls i<<1
#define rs ls | 1
#define mid ((ll+rr)>>1)
#define pii pair<int,int>
#define MP make_pair
typedef long long LL;
const long long INF = 1e18+1LL;
const double Pi = acos(-1.0);
const int N = 5e5+10, maxn = 1e3+20, mod = 1e9+7, inf = 2e9;

LL a,b,p[N],ans1,ans2;
int ok;

int check(LL a,LL B,LL gc) {
            if(a*a - 4*B < 0) return 0;
            LL tmp = (int)(sqrt(a*a - 4*B)+0.00001);
            if(tmp*tmp != a*a - 4*B) return 0;

            LL fi = a+tmp;
            if(fi>=0&&fi%2==0) fi/=2;
            else fi = -1;

            LL se = a-tmp;
            if(se>=0&&se%2==0) se/=2;
            else se = -1;
            if(fi <= 0 && se <= 0) return 0;
            if(fi > 0) {
                    LL x = a - fi;
                    if((__gcd(fi,x)==gc)&&x * fi == B) {
                        ans1 = fi,ans2 = x;
                        if(ans1>ans2)
                            swap(ans1,ans2);
                        ok = 1;
                        return 1;
                    }
                }
                 if(se > 0) {
                    LL x = a - se;
                    if((__gcd(x,se)==gc)&&x * se == B) {
                        ans1 = se,ans2 = x;
                        if(ans1>ans2)
                            swap(ans1,ans2);
                        ok = 1;
                        return 1;
                    }
                }
                return 0;
}
int main() {
        while(scanf("%I64d%I64d",&a,&b)!=EOF) {
            ok = 0;
            for(int i = 1; i * i <= a; ++i) {
                if(a % i == 0) {
                    if(check(a,b*i,i)) break;
                    if(check(a,b*(a/i),a/i)) break;
                }
            }
            if(ok) printf("%I64d %I64d
",ans1,ans2);
            else puts("No Solution");
        }
}

1006题意：给你一个x，然后你要构造一个数组a，满足∑a = x， 任意的i,j( i != j) a[i] != a[j];问你 最大的 s = a1*a2*a3*......*an是多少；

要使得乘积最大，那么相乘的数越多显然S是越大的。。。。

假设x = 7, 那么我们构造出一个数组   2 3 2，到这里有重复的数了，我们就把最后面的2平分到前面  2,3 -> 3，4

假设x = 8，那么我们构造出一个数组   2 3 3，到这里有重复的数了，我们就把最后面的3平分到前面  2,3 -> 3，4 到这里，还有一个1，我们就分到最后一个数上去 -> 3 5

#include<bits/stdc++.h>
using namespace std;
#pragma comment(linker, "/STACK:102400000,102400000")
#define ls i<<1
#define rs ls | 1
#define mid ((ll+rr)>>1)
#define pii pair<int,int>
#define MP make_pair
typedef long long ll;
const long long INF = 1e18+1LL;
const double Pi = acos(-1.0);
const int N = 5e5+10, maxn = 1e3+20, mod = 1e9+7, inf = 2e9;
ll sum[N];
ll pre[N];
ll quick_pow(ll x,ll y)
{
    ll ans=1;
    while(y)
    {
        if(y&1)ans*=x,ans%=mod;
        y>>=1;
        x*=x;
        x%=mod;
    }
    return ans;
}
int main() {
        sum[1]=2;
        for(int i = 2;i<44725 ; ++i) {
                sum[i]=sum[i-1]+i+1;
        }

    int cnt=44720,T;
    pre[0]=1;
    for(int i=1;i<44725;i++)
        {
            pre[i]=(pre[i-1]*(i+1))%mod;
        }
    scanf("%d",&T);
    while(T--)
    {
        ll x;
        scanf("%lld",&x);
        if(x == 1LL) {
            puts("1");
            continue;
        }
        int pos=upper_bound(sum+1,sum+cnt+1,x)-sum-1;
        ll m=x-sum[pos];
        if(m == pos+1) {
            ll ans=pre[pos];
            ans=(ans*(pos+3))%mod;
            ans=(ans*quick_pow(2,mod-2))%mod;
            printf("%lld
",ans);
            continue;
        }
        ll ans = pre[pos - m];
        if(m!=0) ans = ans * ((pre[pos+1]*quick_pow((pre[pos-m+1]),mod-2))% mod )% mod;
        if(x==1)
            printf("1
");
        else
        printf("%lld
",ans);
    }
    return 0;
}

1007题意：一个n点n边的树，每个树节点上有一种颜色苹果最多有k个不同的颜色，问你有多少条路径至少包含了所有颜色的苹果

树分治，这里要用到状态压缩的一点技巧

#include<bits/stdc++.h>
using namespace std;
#pragma comment(linker, "/STACK:102400000,102400000")
#define ls i<<1
#define rs ls | 1
#define mid ((ll+rr)>>1)
#define pii pair<int,int>
#define MP make_pair
typedef long long LL;
const long long INF = 1e18+1LL;
const double Pi = acos(-1.0);
const int N = 100000+10, maxn = 1e3+20, mod = 1e9+7, inf = 2e9;

LL ans = 0;
int n,k,t,head[N],root,a[N],f[N],vis[N],siz[N],allnode;
LL cnt[N],num[N];
struct edge{
    int to,next;
}e[N * 2];
void add(int u,int v) {e[t].next = head[u]; e[t].to = v;head[u] = t++;}
void getroot(int u,int fa) {
        f[u] = 0;
        siz[u] = 1;
        for(int i = head[u]; i != -1; i = e[i].next) {
            int to = e[i].to;
            if(vis[to] || to == fa) continue;
            getroot(to,u);
            siz[u] += siz[to];
            f[u] = max(f[u],siz[to]);
        }
        f[u] = max(f[u],allnode - siz[u]);
        if(f[u] < f[root]) root = u;
}
void getdeep(int u,int fa,int now) {
    for(int i = head[u]; i != -1; i = e[i].next) {
        int to = e[i].to;
        if(to == fa || vis[to]) continue;
        cnt[now|(1<<a[to])]++;
        num[now|(1<<a[to])]++;
        getdeep(to,u,now|(1<<a[to]));
    }
}
LL cal(int u,int now) {
    for(int i = 0; i < (1<<k); ++i) cnt[i] = 0,num[i] = 0;
    num[now]++;
    cnt[now]++;
    getdeep(u,0,now);
    for(int i = 0; i < k; ++i) {
        for(int j = (1<<k)-1; j >= 0; --j) {
            if(!((1<<i)&j)) cnt[j] += cnt[j|(1<<i)];
        }
    }
    LL ans1 = 0;
    for(int i = 0; i < (1<<k); ++i) {
        ans1 += 1LL*num[i]*cnt[i^((1<<k)-1)];
    }
    return ans1;
}
void work(int u) {
        vis[u] = 1;
        ans += cal(u,1<<a[u]);
        for(int i = head[u]; i != -1; i = e[i].next) {
            int to = e[i].to;
            if(vis[to]) continue;
            ans -= cal(to,(1<<a[u])|(1<<a[to]));
            allnode = siz[to];
            root = 0;
            getroot(to,-1);
            work(root);
        }
}
void init() {
        memset(head,-1,sizeof(head));
        memset(vis,0,sizeof(vis));
        t = 1;
        ans = 0;
}
int main() {
        while(scanf("%d%d",&n,&k)!=EOF) {
            for(int i = 1; i <= n; ++i) scanf("%d",&a[i]),a[i]--;
            init();
            for(int i = 1; i < n; ++i) {
                int u,v;
                scanf("%d%d",&u,&v);
                add(u,v);add(v,u);
            }
            f[0] = inf;
            allnode = n;
            root = 0;
            getroot(1,-1);
            work(root);
            printf("%I64d
",ans);
        }
        return 0;
}

1008题意：k个黑球，1个白球，每次每人只能取一球，先取到红球的人胜利，问先取的人是否为有利，或者是平等

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define pi (4*atan(1.0))
#define eps 1e-14
const int N=2e5+10,M=4e6+10,inf=1e9+10,mod=1e9+7;
const ll INF=1e18+10,MOD=1e9+7;
int main()
{
    int x;
    while(~scanf("%d",&x))
    {
        if(x&1)
            printf("0
");
        else
            printf("1
");
    }
    return 0;
}

1009题意：N个角度，长度为D，求出这n个线段围成的面积

#include<bits/stdc++.h>
using namespace std;
#pragma comment(linker, "/STACK:102400000,102400000")
#define ls i<<1
#define rs ls | 1
#define mid ((ll+rr)>>1)
#define pii pair<int,int>
#define MP make_pair
typedef long long LL;
const long long INF = 1e18+1LL;
const double Pi = acos(-1.0);
const int N = 4e5+10, maxn = 1e3+20, mod = 1e9+7, inf = 2e9;

double d;
int n;
int main() {
        while(scanf("%d%lf",&n,&d)!=EOF) {
                double ans = 0.0;
                double x;
                for(int i = 1; i <= n; ++i) {
                    scanf("%lf",&x);
                    ans += d*d*sin(x/180 * Pi)/2;
                }
                printf("%.3f
",ans);
        }
    return 0;
}