Yandex.Algorithm Online Round 3 Sunday, June 15, 2014

Problem A. Distance 求树中距离恰好为2的结点对的个数

#include<iostream>#include<cstdio>#include<cstring>using namespace std;int edge[100007];int main(){    int t;    while(scanf("%d",&t)!=EOF){        for(int i = 1;i <= t; i++)            edge[i] = 0;        int u,v;        for(int i =1 ;i < t; i++){            scanf("%d%d",&v,&u);            edge[u]++;            edge[v]++;        }        long long ans = 0;        for(int i = 1;i <= t; i++){            ans += (1ll*edge[i]*edge[i]-edge[i])/2;        }        cout<<ans<<endl;    }    return 0;}

Problem B. Science

求长度为n只有01的字符串，包含01{k}0子串个数的期望 

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;struct Node{    double max[107][107];};int k;Node operator *(Node a,Node b){    Node c;    memset(c.max,0,sizeof(c.max));    for(int i = 0;i <= k + 2; i++){        for(int j = 0;j <= k+2; j++){            for(int l = 0;l <= k+2; l++){                c.max[i][j] += a.max[i][l]*b.max[l][j];            }        }    }    return c;}Node x,y,z;int main(){    long long n;    while(cin>>n>>k){        memset(x.max,0,sizeof(x.max));        for(int i = 0;i <= k+2 ;i++)            x.max[i][i] = 1.0;        memset(y.max,0,sizeof(y.max));        for(int i = 0;i <= k; i++)            y.max[0][i] = 0.5;        for(int i = 1;i <= k+1; i++)            y.max[i][i-1] = 0.5;        y.max[0][k+1] = 0.5;        y.max[k+1][k+1] = 0.5;        y.max[k+2][k] = 0.5;        y.max[k+2][k+2] = 1.0;        n--;        while(n){            if(n&1) x = x*y;            y = y * y;            n /= 2;        }        double ans = x.max[k+2][0]*0.5 + x.max[k+2][k+1]*0.5;        printf("%0.16f\n",ans);    }    return 0;}

Problem C. Intervals

求有几个区间，在这个区间中存在两个数，差值为d

#include<iostream>#include<cstring>#include<algorithm>#include<set>#include<map>#include<cstdio>#define ll long longusing namespace std;map<ll,int> haha;int main(){    int n,d;    while(scanf("%d%d",&n,&d)!=EOF){        haha.clear();        int be = 0;        haha[0] = 0;        ll sum = 0,u;        ll ans = 0;        for(int i = 1;i <= n; i++){            scanf("%I64d",&u);            if(haha.find(u-d)!= haha.end())                be = max(be,haha[u-d]);            if(haha.find(u+d) != haha.end())                be = max(be,haha[u+d]);            ans += be;            haha[u] = i;        }        cout<<ans<<endl;    }    return 0;}

F. Weird Game

对于长度为N的一维格子，每次可以从中选出恰好长度为L，且不包含被染色位置的格子。

如果找不到可以染色的区间了。算输。

sg函数 对于可以选择长度为L进行染色，算出 格子长度为1-7000的sg函数。

由于对于一个N，要枚举分出的两个线段长度，需要N的转移，所以是N*N的复杂度，会超时。

打表找规律，发现大于8的L的必败态有固定的增长速度。于是，打表。

小于8的就直接sg函数做了就行。

#include<cstdio>#include<cstring>#include<algorithm>#include<bitset>#include<iostream>#include<set>using namespace std;int check[7000];int sg[7000];set<int> haha[70001];int change[14][2]={2,0,2,0,4,-2,4,-2,4,0,4,-2,8,-2,4,-2,8,0,8,-2,16,-6,4,0};int main(){    int cnt = 1;    for(int i = 2;i <= 7;i++){        memset(sg,0,sizeof(sg));        for(int j = i; j <= 7000; j++){            int u = j - i;            for(int k = 0;k + k <= u; k++){                check[sg[k]^sg[u-k]] = cnt;            }            for(int k = 0;;k++){                if(check[k] != cnt){                    sg[j] = k;                    break;                }            }            cnt++;        }        int f = 0;        for(int j = i;j <= 7000; j++){           if(sg[j] == 0){               haha[i].insert(j);           }        }    }    for(int i = 8; i<=7000;i++){        int j = i -1;        for(int l = 0;l < 12; l++){            j = j + change[l][0]*i+change[l][1];            if(j > 7000) break;            haha[i].insert(j);        }    }    int n;    while(scanf("%d",&n)!=EOF){        if(n % 2 == 0) printf("S");        else printf("F");        for(int i = 2;i <= n; i++){            if(haha[i].find(n) != haha[i].end()) printf("S");            else printf("F");        }        printf("\n");    }    return 0;}