第八届福建省大学生程序设计竞赛训练总结【7/12】

训练结果：Rank 32 

Ac题数：6

tot time：1049

A.水题。鸡兔同笼问题，保证有解，小学六年级问题。

被I64d卡了三发= =.

B.计算几何，窝不会，队长会丫。

队长Ac代码：

#include<stdio.h>#include<string.h>#include<math.h>using namespace std;#define LL long long intconst int N = 1e5 + 7;const int MOD = 1e9+7;/**********************************/struct point{    double x,y;};struct trangle{    point p[3];}c[2];double multi(point p1,point p2,point p0){    return fabs((p1.x-p0.x)*(p2.y-p0.y)-(p1.y-p0.y)*(p2.x-p0.x));}double area(point a,point b,point c){    return multi(a,b,c);}int sum;// 判断c1 在不在c2 里bool cantian(trangle c1,trangle c2){    double ac2 =area(c2.p[0],c2.p[1],c2.p[2]);    int t =0 ;    for(int i=0;i<=2;i++)        if(area(c1.p[i],c2.p[0],c2.p[1])+area(c1.p[i],c2.p[1],c2.p[2])+           area(c1.p[i],c2.p[0],c2.p[2]) > ac2);        else sum++,t++;    return t==3;}void solve(){    sum = 0;    if(cantian(c[0],c[1])||cantian(c[1],c[0])){        puts("contain");        return ;    }    if(sum == 0){        puts("disjoint");        return ;    }    puts("intersect");    return ;}int main(){    int _ = 1,kcase = 0;    for(scanf("%d",&_);_--;){        for(int i=0;i<=1;i++)for(int j=0;j<=2;j++)            scanf("%lf%lf",&c[i].p[j].x,&c[i].p[j].y);        solve();    }    return 0;}

C.

D.一道挺好的思维KMP的题目，我是萌萌哒D题题解

E.

F.窝不会啊，队长会啊，贼强啊。

队长Ac代码：

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;#define LL long long intint n;const int N=3e5+7;const int MOD=1e9+7;struct edge{    int to,next;}G[N<<1];int head[N],cntG;void add(int u,int v){    G[cntG].to=v,G[cntG].next=head[u],head[u]=cntG++;    G[cntG].to=u,G[cntG].next=head[v],head[v]=cntG++;}LL bit[N][3];#define lowbit(x) (x&-x)void update(int i,int v,int x){    for(;i&&i<=n;i+=lowbit(i)){        bit[i][x]+=v; bit[i][x]%=MOD;    }}LL getSum(int i,int x){    LL ans = 0;    for(;i;i-=lowbit(i))        ans+=bit[i][x];    return ans%=MOD;}int dep[N],fa[N],son[N],sz[N];void dfs(int u,int f,int d){    dep[u]=d,fa[u]=f,son[u]=0,sz[u]=1;    for(int i=head[u],to;i!=-1;i=G[i].next){        to=G[i].to;        if(to == f) continue;        dfs(to,u,d+1);        sz[u]+=sz[to];        if(sz[to]>sz[son[u]]) son[u]=to;    }}int top[N],tree[N],tot;void dfs2(int u,int tp){    tree[u]=++tot;top[u]=tp;    if(son[u]) dfs2(son[u],tp);    else return ;    for(int i=head[u],to;i!=-1;i=G[i].next){        to=G[i].to;        if(to==fa[u]||to==son[u]) continue;        dfs2(to,to);    }}void solve(int x){    int fx=top[x],t=dep[x];    LL sum1 = 0,sum2 = 0,sum3 = 0;    while(fx!=1){        sum1 = (sum1+getSum(tree[x],0)-getSum(tree[fx]-1,0)+MOD)%MOD;        sum2 = (sum2+getSum(tree[x],1)-getSum(tree[fx]-1,1)+MOD)%MOD;        sum3 = (sum3+getSum(tree[x],2)-getSum(tree[fx]-1,2)+MOD)%MOD;        x=fa[fx],fx=top[x];    }    sum1 = (sum1+(getSum(tree[x],0)-getSum(tree[1]-1,0))+MOD)%MOD;    sum2 = (sum2+(getSum(tree[x],1)-getSum(tree[1]-1,1))+MOD)%MOD;    sum3 = (sum3+(getSum(tree[x],2)-getSum(tree[1]-1,2))+MOD)%MOD;    sum1*=t;sum1%=MOD;    sum2-=sum1;sum2=(sum2%MOD+MOD)%MOD;    sum2+=sum3;(sum2%=MOD);    printf("%I64d\n",sum2);    return ;}int read(){    int res = 0, ch, flag = 0;    if((ch = getchar()) == '-')             //判断正负        flag = 1;    else if(ch >= '0' && ch <= '9')           //得到完整的数        res = ch - '0';    while((ch = getchar()) >= '0' && ch <= '9' )        res = res * 10 + ch - '0';    return flag ? -res : res;}int main(){    int _ = 1,kcase = 0;    for(scanf("%d",&_);_--;){        memset(head,-1,sizeof(head));        memset(bit,0,sizeof(bit));        n=read();cntG=0;        for(int i=2,x;i<=n;i++){            x=read();add(i,x);        }        dfs(1,0,1);        tot=0,dfs2(1,1);        int q,op,v,x,k;        for(q=read();q--;){            op=read(),v=read();            if(1 == op){                x=read(),k=read();                update(tree[v],k,0);                update(tree[v],(LL)k*dep[v]%MOD,1);                update(tree[v],x,2);            }            else solve(v);        }    }    return 0;}

G.一道很好的期望思维题，将整体拆分成若干个独立事件去求期望，求和的一种解题思路，我是萌萌哒G题题解

H.

I.

J.

K.错排+Cnm，不难。我是萌萌哒K题题解

L.爆搜即可。我是萌萌哒L题题解

队长Ac代码：

#include<stdio.h>#include<string.h>#include<math.h>using namespace std;#define LL long long intconst int N = 1e5 + 7;const int MOD = 1e9+7;/**********************************/int a[10][10];bool judge(int x){    if(x==a[1][1]&&a[1][1]==a[1][2]&&a[1][2]==a[1][3])return true;    if(x==a[2][1]&&a[2][1]==a[2][2]&&a[2][2]==a[2][3])return true;    if(x==a[3][1]&&a[3][1]==a[3][2]&&a[3][2]==a[3][3])return true;    if(x==a[1][1]&&a[1][1]==a[2][1]&&a[2][1]==a[3][1])return true;    if(x==a[1][2]&&a[1][2]==a[2][2]&&a[2][2]==a[3][2])return true;    if(x==a[1][3]&&a[1][3]==a[2][3]&&a[2][3]==a[3][3])return true;    if(x==a[1][1]&&a[1][1]==a[2][2]&&a[2][2]==a[3][3])return true;    if(x==a[1][3]&&a[1][3]==a[2][2]&&a[2][2]==a[3][1])return true;    return false;}bool Kim2(int x){    int flag = 0;    for(int i=1;i<=3;i++){        for(int j=1;j<=3;j++){            if(a[i][j]==0){                a[i][j]=x;                if(judge(x)) flag++;                a[i][j]=0;            }        }    }    return flag>=2;}bool Kim1(int x){    for(int i=1;i<=3;i++){        for(int j=1;j<=3;j++){            if(a[i][j]==0){                a[i][j]=x;                if(judge(x)||Kim2(x))                    return true;                a[i][j]=0;            }        }    }    return false;}char s[10];int main(){    int _ = 1,kcase = 0;    for(scanf("%d",&_);_--;){        for(int i=1;i<=3;i++){            for(int j=1;j<=3;j++){                scanf("%s",s);                if(s[0]=='.') a[i][j]=0;                else if(s[0]=='x') a[i][j]=1;                else a[i][j]=2;            }        }        scanf("%s",s);int x;        if(s[0]=='x')   x=1;        else            x=2;        if(judge(3-x)) puts("Cannot win!");        else if(judge(x)||Kim1(x))             puts("Kim win!");        else puts("Cannot win!");    }    return 0;}