9.10+9.14+9.16

9.10
40+0+60=100 rank 16
T1 裸的exgcd，然而不会求解的个数了，用解析几何搞的，考试时一堆问题都没调出来。。。
T2树形dp，f[i][j]表示i这颗子树里选j个黑点的最大收益，像背包一样转移就好了，考试打的暴力，还tm翻车了(0x7f)
T3神题，至今没改，留个坑以后填吧。。。
T1

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<cmath>
#define LL long long
using namespace std;
LL exgcd(LL a,LL b,LL &x,LL &y){
    if(b==0){
        x=1;y=0;
        return a;
    }
    LL g=exgcd(b,a%b,x,y);
    LL t=x;
    x=y;
    y=t-(a/b)*x;
    return g;
}
int main(){
    //freopen("fuction.in","r",stdin);
    //freopen("fuction.out","w",stdout);
    LL a,b,c,d,x,y,ans,numx,numy,T;
    scanf("%lld",&T);
    while(T--){
        scanf("%lld%lld%lld",&a,&b,&c);
        if(a==0&&b==0){
            printf("0
");
            continue;
        }
        if(a==0){
            if((c%b==0)&&(c/b>0)) printf("ZenMeZheMeDuo
");
            else printf("0
");
            continue;
        }
        if(b==0){
            if((c%a==0)&&((c/a>0))) printf("ZenMeZheMeDuo
");
            else  printf("0
");
            continue;
        }
        if(a<0||b<0){a=-a;b=-b;c=-c;}
        LL d=exgcd(a,b,x,y);
        if(c%d!=0){
            printf("0
");
            continue;
        }
        if(a*b<0){
            printf("ZenMeZheMeDuo
");
            continue;
        }
        x=x*(c/d);y=y*(c/d);
        if(x<=0&&y<=0){
            printf("0
");
            continue;
        }
        if(x<=0&&y>0){
            numy=(y-1)/(a/d)+1;
            if(x==0)numx=1;
            else numx=(-x)/(b/d)+1;
            ans=numy-numx;
            if(ans<0)ans=0;
            if(ans>65535)  printf("ZenMeZheMeDuo
");
            else printf("%lld
",ans);
            continue;
        }
        if(y<=0&&x>0){
            numx=(x-1)/(b/d)+1;
            if(y==0) numy=1;
            else numy=(-y)/(a/d)+1;
            ans=numx-numy;
            if(ans<0)ans=0;
            if(ans>65535)  printf("ZenMeZheMeDuo
");
            else printf("%lld
",ans);
            continue;
        }
        if(x>0&&y>0){
            numx=(x-1)/(b/d)+1;
            numy=(y-1)/(a/d)+1;
            ans=numx+numy-1;
            if(ans<0)ans=0;
            if(ans>65535)  printf("ZenMeZheMeDuo
");
            else printf("%lld
",ans);
            continue;
        }
    }
    return 0;
}

T2

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<cmath>
#define N 2050
using namespace std;
int e=1,head[N];
struct edge{
    int u,v,w,next;
}ed[2*N];
void add(int u,int v,int w){
    ed[e].u=u;ed[e].v=v;ed[e].w=w;
    ed[e].next=head[u];head[u]=e++;
}
int n,m;
int size[N],fa[N];
long long f[N][N];
void dfs(int x){
    size[x]++;
    for(int i=head[x];i;i=ed[i].next){
        int v=ed[i].v;
        if(v==fa[x])continue;
        fa[v]=x; dfs(v);
        size[x]+=size[v];
        for(int j=min(size[x],m);j>=0;j--){
            int minn=max(0,max(j-size[x]+size[v],size[v]-n+m));
            for(int k=minn;k<=min(size[v],min(m,j));k++){
                long long tmp=(long long)(k*(m-k)+(size[v]-k)*((n-m)-(size[v]-k)));
                f[x][j]=max(f[x][j],f[x][j-k]+f[v][k]+(ed[i].w*tmp));
            }
        }
    }
}
int main(){
    scanf("%d%d",&n,&m);
    int u,v,w;
    for(int i=1;i<n;i++){
        scanf("%d%d%d",&u,&v,&w);
        add(u,v,w); add(v,u,w);
    }
    dfs(1);
    printf("%lld
",f[1][m]);
    return 0;
}

9.14
70+70+15=155 rank1
T1 欧拉路，去掉的至少有一个自环或者两条边有公共点(欧拉路需要保证整个图只有0或2个有奇数条连边)需要判边是否连通！最后手贱还扔了10分……
T2 下底分块的思想 我们要求一个最大的d，满足

∑i=1n(⌈ai/d⌉∗d−ai)≤k

设C=k+∑ni=1ai
则∑ni=1⌈ai/d⌉≤⌊C/d⌋ 易证等式左边随d增加单调不增，所以按右边分块，可能作为答案的是所有块的右端点，考试打了好多特判骗了70
T3 神奇的dp….dp[i][j]表示一棵i-超级树，同时有j条点不重复的路径存在的方案数
枚举左子树和右子树的路径条数l、r，记num=dp[i][l]*dp[i][r]，
什么也不做 dp[i+1][l+r]+=num
根自己作为一条新路径 dp[i+1][l+r+1]+=num
根连接到左子树（或右子树）的某条路径上 dp[i+1][l+r]+=2*num*(l+r)
根连接左子树和右子树的各一条路径 dp[i+1][l+r-1]+=2*num*l*r
根连接左子树（或右子树）的两条路径 dp[i+1][l+r-1]+=num*(l*(l-1)+r*(r-
1))
注意这里的路径都是有向的！
T1

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<cmath>
using namespace std;
long long ans,in[100005],tot,n,m;
int fa[100005];
bool vis[100005];
int find(int x){
    if(x==fa[x])return x;
    return fa[x]=find(fa[x]);
}
int main(){
    scanf("%lld%lld",&n,&m);
    int u,v,fu,fv;
    for(int i=1;i<=n;i++)fa[i]=i;
    for(int i=1;i<=m;i++){
        scanf("%d%d",&u,&v);
        vis[u]=1;vis[v]=1;
        if(u!=v){
            in[u]++;in[v]++;
            fu=find(u);fv=find(v);
            if(fu!=fv)fa[fv]=fu;
        }
        else tot++;
    }
    for(int i=1;i<=n;i++)if(vis[i]){fu=find(i);break;}
    for(int i=1;i<=n;i++)if(vis[i]&&find(i)!=fu){printf("0
");return 0;}
    for(int i=1;i<=n;i++)
        ans+=in[i]*(in[i]-1)/2;
    ans+=tot*(m-tot);
    ans+=tot*(tot-1)/2;
    printf("%lld
",ans);
    return 0;
}

T2

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<cmath>
#define LL long long
#define N 105
using namespace std;
int n;
LL m,a[N],tot; 
bool check(LL x){
    LL cnt=0,top=tot/x;
    for(int i=1;i<=n;++i){
        cnt+=((a[i]-1)/x)+1;
        if(cnt>top)return 0;
    }return 1;
}
int main(){
    scanf("%d%lld",&n,&m);
    for(int i=1;i<=n;i++){
        scanf("%lld",&a[i]);
        tot+=a[i];
    }
    tot+=m;
    for(LL d=tot;d>=1;--d){
        if(check(d)){printf("%lld
",d);return 0;}
        d=(tot/(tot/d+1))+1;
    }
    return 0;
}

T3

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<cmath>
#define LL long long
#define N 302
using namespace std;
int n;
LL f[N][N],num,mod;
int main(){
    scanf("%d%lld",&n,&mod);
    f[1][0]=1; f[1][1]=1;
    for(int i=1;i<n;i++)
        for(int l=0;l<=n-i+1;l++)
            for(int r=0;r+l<=n-i+2;r++){
                num=f[i][l]*f[i][r]%mod;
                (f[i+1][l+r]+=num+2*num*(l+r))%=mod;
                (f[i+1][l+r+1]+=num)%=mod;
                if(l+r<=0)continue;
                if(l==0) (f[i+1][l+r-1]+=num*(2*l*r+r*(r-1)))%=mod;
                if(r==0) (f[i+1][l+r-1]+=num*(2*l*r+l*(l-1)))%=mod;
                if(l!=0&&r!=0) (f[i+1][l+r-1]+=num*(2*l*r+l*(l-1)+r*(r-1)))%=mod;
            }
    printf("%lld
",f[n][1]%mod);
    return 0;
}

9.16
100+100+40=240 rank3
T1 贪心找能剩下的最多的
T2 ans=Cn2∗n
T3 竟然是区间dp，二维树状数组也可以水！想到回文那一串东西无法自拔，关键是我什么都不会……看了就要学，不学不要看。。。。
f[i][j]=f[i+1][j]+f[i][j-1]-f[i+1][j-1]+[i~j是回文串]
T1

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<cmath>
#define N 100005
#define LL long long
using namespace std;
struct data{LL a,b,c;}d[N];
bool cmpc(data a,data b){return a.c>b.c;}
LL n,tot,ans;
LL read(){
    LL a=0;char ch=getchar();
    while(ch<'0'||ch>'9')ch=getchar();
    while(ch>='0'&&ch<='9'){a=a*10+ch-'0';ch=getchar();}
    return a;
}
int T;
int main(){
    scanf("%d",&T);
    while(T--){
        tot=0; ans=0; n=read();
        for(int i=1;i<=n;i++){
            d[i].a=read();
            d[i].b=read();
            d[i].c=d[i].b-d[i].a;
        }
        sort(d+1,d+n+1,cmpc);
        for(int i=1;i<=n;i++){
            if(tot>=d[i].b)tot-=d[i].a;
            else{ans+=d[i].b-tot;tot=d[i].c;}
        }
        printf("%lld
",ans);
    }
    return 0;
}

T2

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<cmath>
#define LL long long
#define N 1000005
#define mod 1000000007
using namespace std;
LL fac[2*N],n,T,ans;
LL qp(LL a,LL b){
    LL ff=1;
    while(b){
        if(b&1)ff=ff*a%mod;
        a=a*a%mod;b>>=1;
    }
    return ff;
}
LL C(LL x,LL y){
    if(!y||y==x)return 1;
    return ((fac[x]*qp(fac[y],mod-2))%mod*qp(fac[x-y],mod-2))%mod;
}
int main(){
    fac[1]=1;
    for(int i=2;i<=2000000;i++)fac[i]=(fac[i-1]*i)%mod;
    scanf("%lld",&T);
    while(T--){
        scanf("%lld",&n);
        printf("%lld
",C(2*n,n));
    }
    return 0;
}

T3

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<cmath>
#define N 5005
using namespace std;
char s[N];
int len,f[N][N],g[N][N];
int read(){
    int a=0;char ch=getchar();
    while(ch<'0'||ch>'9')ch=getchar();
    while(ch>='0'&&ch<='9'){a=a*10+(ch^48);ch=getchar();}
    return a;
}
int main(){
    scanf("%s",s+1);
    len=strlen(s+1);
    for(int i=1;i<=len;++i){
        g[i][i]=1;
        for(int j=1;;++j){
            if(i+j>len||j>=i) break;
            if(s[i+j]==s[i-j])g[i-j][i+j]=1;
            else break;
        }
    }
    for(int i=1;i<=len;++i){
        for(int j=1;;++j){
            if(i+j>len||j>i) break;
            if(s[i+j]==s[i-j+1])g[i-j+1][i+j]=1;
            else break;
        }
    }
    for(int i=1;i<=len;i++)f[i][i]=1;
    for(int l=2;l<=len;l++){
        for(int i=1;i<=len-l+1;i++){
            int j=i+l-1;
            f[i][j]=f[i][j-1]+f[i+1][j]-f[i+1][j-1]+g[i][j];
        }
    }
    int l,r,T=read();
    while(T--){
        l=read();r=read();
        printf("%d
",f[l][r]);
    }
    return 0;
}