2017-10-6 清北刷题冲刺班a.m

1.角谷猜想

#include<iostream>
#include<cstdio>
#include<cstring>
#define maxn 10010
using namespace std;
char ch[maxn],st[maxn];
int t,top;
int main(){
    //freopen("Cola.txt","r",stdin);
    freopen("kakutani.in","r",stdin);freopen("kakutani.out","w",stdout);
    scanf("%d",&t);
    while(t--){
        scanf("%s",ch+);
        int len=strlen(ch+);
        top=;
        for(int i=;i<=len;i++){
            if(top>&&st[top]==''&&ch[i]==''){//出现13
                top=top-;
            }
            else if(ch[i]!=''&&ch[i]!=''){
                top=top+;
                st[top]=ch[i];
            }
        }
        if(top==){
            printf("0\n");
            continue;
        }
        for(int i=;i<=top;i++){
            printf("%c",st[i]);
        }
        printf("\n");
    }
    fclose(stdin);fclose(stdout);
    return ;
}

100分 栈模拟

2.刀塔（二分答案）

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define N 1000010
using namespace std;
int x,y,Asd,jk,A,B,k,ans,t,q,n,Max,nm,a[N],f[N][],Min[N];
void init(){
    Min[]=-;
    for(int i=;i<=n;i++){
        Min[i]=((i&(i-))==)?Min[i-]+:Min[i-];
        f[i][]=a[i];
    }
    for(int j=;j<=Min[n];j++)
        for(int i=;i+(<<j)-<=n;i++)
            f[i][j]=min(f[i][j-],f[i+(<<(j-))][j-]);
}
int find(int L,int R){
    int k=Min[R-L+];
    return min(f[L][k],f[R-(<<k)+][k]);
}
int main(){
    freopen("dota.in","r",stdin);freopen("dota.out","w",stdout);
    scanf("%d%d%d%d",&n,&A,&B,&k);
    for(int i=;i<=n;i++)
        scanf("%d",&a[i]);
    init();
    for(int i=;i<=n-B-k+;i++){
        Max=B-A;
        for(int j=A;j<=Max;j++){
            nm=B-j;
            Asd=find(i,i+j-);
            jk=find(i+j+k,i+j+k+nm-);
            ans=max(ans,min(Asd,jk));
        }
    }
    cout<<ans;
}

50分 倍增表暴力

/*
    题目可以简化为选择一段长为B+K的区间，中间去掉k个数，且去掉k个数后两侧的数都多于a
    二分答案，对于每个答案，check的方法就是枚举中间区间的位置，枚举两侧区间最大满足答案的长度判断是否满足上面的条件即可
*/
#include<cstdio>
#include<algorithm>
#define maxn 2000010
using namespace std;
int x[maxn],l[maxn],r[maxn];
int n,a,b,k,ans;
bool check(int c){
    for(int i=;i<=n;i++){
        if(x[i]>=c)l[i]=l[i-]+;
        else l[i]=;
    }
    for(int i=n;i>=;i--){
        if(x[i]>=c)r[i]=r[i+]+;
        else r[i]=;
    }
    for(int i=;i<=n;i++){
        if(l[i-]>=a&&r[i+k]>=a&&l[i-]+r[i+k]>=b)return ;
    }
    return ;
}
int main(){
    freopen("dota.in","r",stdin);freopen("dota.out","w",stdout);
    //freopen("dota10.in","r",stdin);
    int left,right,mid;
    scanf("%d%d%d%d",&n,&a,&b,&k);
    for(int i=;i<=n;i++){
        scanf("%d",&x[i]);
        if(x[i]>right)right=x[i];
    }
    left=,right=right+;
    while(left<=right){
        int mid=(left+right)>>;
        if(check(mid))ans=mid,left=mid+;
        else right=mid-;
    }
    printf("%d",ans);
}

100分 二分答案

3.反击数（kmp+数位dp）

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
char p[];
int m,f[],a[];
long long dp[][][];
long long l,r,k;
void KMP(){
    f[]=;f[]=;
    for(int i=;i<m;i++){
        int j=f[i];
        while(j&&p[i]!=p[j])j=f[j];
        f[i+]=p[i]==p[j]?j+:;
    }
}
long long dfs(int k,int w,bool limit,bool get){
    if(!k)return get;
    else if (!limit&&dp[k][w][get]!=-)return dp[k][w][get];
    else {
        long long ans=;
        for(int i=,j=limit?a[k]:;i<=j;i++){
            int t=w;
            while(t&&p[t]-''!=i)t=f[t];
            if(p[t]-''==i)t++;
            ans+=dfs(k-,t,limit&&(i==a[k]),get||(t==m));
        }
        return limit?ans:dp[k][w][get]=ans;
    }
}
long long query(long long x){
    int cnt=;
    while(x){
        a[++cnt]=x%;
        x/=;
    }
    memset(dp,-,sizeof(dp));
    return dfs(cnt,,,);
}
int main(){
    freopen("spenum.in","r",stdin);freopen("spenum.out","w",stdout);
    //freopen("Cola.txt","r",stdin);
    cin>>l>>r;scanf("%s",p);cin>>k;
    m=strlen(p);
    KMP();
    k+=query(l-);
    if(query(r-)<k){
        printf("Hey,wake up!");
        return ;
    }
    long long tmp=,ans=l-;
    while(tmp<r-l+)tmp=tmp*;
    while(tmp){
        if(query(ans+tmp)<k)ans+=tmp;
        tmp=tmp/;
    }
    cout<<ans+;
    return ;
}

100分 kmp+数位dp