2017-10-6 清北刷题冲刺班a.m

Posted 2020-10-09 Soda

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+1);
        int len=strlen(ch+1);
        top=0;
        for(int i=1;i<=len;i++){
            if(top>0&&st[top]==\'1\'&&ch[i]==\'3\'){//出现13 
                top=top-1;
            }
            else if(ch[i]!=\'7\'&&ch[i]!=\'4\'){
                top=top+1;
                st[top]=ch[i];
            }
        }
        if(top==0){
            printf("0\\n");
            continue;
        }
        for(int i=1;i<=top;i++){
            printf("%c",st[i]);
        }
        printf("\\n");
    }
    fclose(stdin);fclose(stdout);
    return 0;
}

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][20],Min[N];
void init(){
    Min[0]=-1;
    for(int i=1;i<=n;i++){
        Min[i]=((i&(i-1))==0)?Min[i-1]+1:Min[i-1];
        f[i][0]=a[i];
    }
    for(int j=1;j<=Min[n];j++)
        for(int i=1;i+(1<<j)-1<=n;i++)
            f[i][j]=min(f[i][j-1],f[i+(1<<(j-1))][j-1]);
}
int find(int L,int R){
    int k=Min[R-L+1];
    return min(f[L][k],f[R-(1<<k)+1][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=1;i<=n;i++)
        scanf("%d",&a[i]);
    init();
    for(int i=1;i<=n-B-k+1;i++){
        Max=B-A;
        for(int j=A;j<=Max;j++){
            nm=B-j;
            Asd=find(i,i+j-1);
            jk=find(i+j+k,i+j+k+nm-1);
            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=1;i<=n;i++){
        if(x[i]>=c)l[i]=l[i-1]+1;
        else l[i]=0;
    }
    for(int i=n;i>=1;i--){
        if(x[i]>=c)r[i]=r[i+1]+1;
        else r[i]=0;
    }
    for(int i=1;i<=n;i++){
        if(l[i-1]>=a&&r[i+k]>=a&&l[i-1]+r[i+k]>=b)return 1;
    }
    return 0;
}
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=1;i<=n;i++){
        scanf("%d",&x[i]);
        if(x[i]>right)right=x[i];
    }
    left=1,right=right+1;
    while(left<=right){
        int mid=(left+right)>>1;
        if(check(mid))ans=mid,left=mid+1;
        else right=mid-1;
    }
    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[32];
int m,f[32],a[32];
long long dp[20][2000][2];
long long l,r,k;
void KMP(){
    f[0]=0;f[1]=0;
    for(int i=1;i<m;i++){
        int j=f[i];
        while(j&&p[i]!=p[j])j=f[j];
        f[i+1]=p[i]==p[j]?j+1:0;
    }
}
long long dfs(int k,int w,bool limit,bool get){
    if(!k)return get;
    else if (!limit&&dp[k][w][get]!=-1)return dp[k][w][get];
    else {
        long long ans=0;
        for(int i=0,j=limit?a[k]:9;i<=j;i++){
            int t=w;
            while(t&&p[t]-\'0\'!=i)t=f[t];
            if(p[t]-\'0\'==i)t++;
            ans+=dfs(k-1,t,limit&&(i==a[k]),get||(t==m));
        }
        return limit?ans:dp[k][w][get]=ans;
    }
}
long long query(long long x){
    int cnt=0;
    while(x){
        a[++cnt]=x%10;
        x/=10;
    }
    memset(dp,-1,sizeof(dp));
    return dfs(cnt,0,1,0);
}
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-1);
    if(query(r-1)<k){
        printf("Hey,wake up!");
        return 0;
    }
    long long tmp=1,ans=l-1;
    while(tmp<r-l+1)tmp=tmp*2;
    while(tmp){
        if(query(ans+tmp)<k)ans+=tmp;
        tmp=tmp/2;
    }
    cout<<ans+1;
    return 0;
}

100分 kmp+数位dp