牛客练习赛79

牛客练习赛79

A. 炼金术师
单调栈裸题，很容易想到对于每一次的粉刷，都要找到他上一次粉刷到的比他大的区间（就是单调栈中栈顶的下一个元素），在他的基础上加一种颜色。最后的结果只是判一下单调栈中最后存在多少个数。

代码

#include <bits/stdc++.h>
#define ll long long
#define mem( f, x ) memset( f, x, sizeof(f))
#define pii pair<int,int>
#define fi first
#define se second
#define mk(x,y) make_pair(x,y)
#define pk push_back

using namespace std;
const int M = 1e6+5;
const int N = 50;
const int MOD = 1e9+7;
const int INF = 0x3f3f3f3f;
int m, n;
int a[N], cnt[N];
int sk[N], ct;

int main( ){
    while( scanf( "%d", &n ) != EOF ){
        ct = 0;
        for( int i = 0; i < n; i++ ){
            scanf( "%d", a+i );
            while( ct && a[sk[ct-1]] <= a[i] )
                ct--;
            if( !ct )
                cnt[i] = 1;
            else
                cnt[i] = cnt[sk[ct-1]]+1;
            sk[ct++] = i;
        }
        printf( "%d\n", cnt[n-1] );
    }
    return 0;
}

B. 刀工对决
要将两根面条切成一样长，有两种刀，用BFS处理出切的过程中达到的长度，两个面条切的过程中相遇的时候就是最少的切的次数。

代码

#include <bits/stdc++.h>
#define ll long long
#define mem( f, x ) memset( f, x, sizeof(f))
#define pii pair<int,int>
#define fi first
#define se second
#define mk(x,y) make_pair(x,y)
#define pk push_back

using namespace std;
const int M = 1e6+5;
const int N = 50;
const int MOD = 1e9+7;
const int INF = 0x3f3f3f3f;
int m, n;
unordered_map<int,int> mpx;
unordered_map<int,int> mpy;

void Bfs1( int x ){
    queue<int> q;
    mpx.clear();
    q.push( x );
    mpx[x] = 0;
    int cur, nxt;
    while( !q.empty() ){
        cur = q.front();
        q.pop();
        if( cur%3 == 0 ){
            nxt = cur/3;
            if( !mpx.count(nxt) )
                mpx[nxt] = mpx[cur]+1;
            else{
                if( mpx[nxt] > mpx[cur]+1 )
                    mpx[nxt] = mpx[cur]+1;
            }
            q.push( nxt );
        }
        if( cur%5 == 0 ){
            nxt = cur/5*3;
            if( !mpx.count(nxt) )
                mpx[nxt] = mpx[cur]+1;
            else{
                if( mpx[nxt] > mpx[cur]+1 )
                    mpx[nxt] = mpx[cur]+1;
            }
            q.push( nxt );
        }
    }
}

int Bfs2( int y ){
    queue<int> q;
    mpy.clear();
    q.push( y );
    mpy[y] = 0;
    int cur, nxt;
    while( !q.empty() ){
        cur = q.front();
        q.pop();
        if( mpx.count( cur ) )
            return mpy[cur]+mpx[cur];
        if( cur%3 == 0 ){
            nxt = cur/3;
            if( !mpy.count(nxt) )
                mpy[nxt] = mpy[cur]+1;
            else{
                if( mpy[nxt] > mpy[cur]+1 )
                    mpy[nxt] = mpy[cur]+1;
            }
            q.push( nxt );
        }
        if( cur%5 == 0 ){
            nxt = cur/5*3;
            if( !mpy.count(nxt) )
                mpy[nxt] = mpy[cur]+1;
            else{
                if( mpy[nxt] > mpy[cur]+1 )
                    mpy[nxt] = mpy[cur]+1;
            }
            q.push( nxt );
        }
    }
    return -1;
}

int main( ){
    while( scanf( "%d", &n ) != EOF ){
        int x, y;
        int ans = 0;
        bool flag = 0;
        for( int i = 0; i < n; i++ ){
            scanf( "%d %d", &x, &y );
            if( !flag ){
                Bfs1( x );
                int tmp = Bfs2( y );
                if(tmp == -1 )
                    flag = 1;
                ans += tmp;
            }
        }
        if( !flag )
            printf( "%d\n", ans );
        else
            printf("-1\n" );
    }
    return 0;
}

C. 小G的GCD
牛客套路题，凡是给了代码，先粘进编译器跑一发。我是打表打到n=20的时候，确定了是个斐波那契的变形（不会证明）。最终的答案就是首项为1，次项为2的斐波那契数列的下标加2，在这个斐波那契数列中找到最大且小于n的项即可。

代码

#include <bits/stdc++.h>
#define ll long long
#define mem( f, x ) memset( f, x, sizeof(f))
#define pii pair<int,int>
#define fi first
#define se second
#define mk(x,y) make_pair(x,y)
#define pk push_back

using namespace std;
const int M = 105;
const int N = 1e5+5;
const int MOD = 1e9+7;
const int INF = 0x3f3f3f3f;
ll m, n;
ll f[100];
int cnt;
/*
long long GCD(long long x, long long y) {
     if (!y) return 1ll;
     return GCD(y, x % y) + 1ll;
}

long long maxGCD(long long n) {
     long long ans = 0ll;
     ll x, y;
     for (long long i = 1; i <= n; i++)
         for (long long j = 1; j <= n; j++){
            ll tmp = GCD( i, j );
            if( ans < tmp ){
                x = i, y = j;
                ans = tmp;
            }
         }
    printf( "%lld %lld\n", x, y );
     return ans;
 }



 int main( ){
    while( scanf( "%lld", &n ) != EOF ){
        cout << maxGCD( n ) << endl;
    }
    return 0;
 }
*/

void gao( ){
    f[0] = 1, f[1] = 2;
    int i = 2;
    ll nxt = f[0]+f[1];
    cnt = 2;
    while( nxt <= 1e18 ){
        f[cnt++] = nxt;
        nxt = f[cnt-1]+f[cnt-2];
    }
 }

 int main( ){
    gao( );
    while( scanf( "%lld", &n ) != EOF ){
        int pos = 0;
        while( f[pos] <= n )
            pos++;
        pos--;
        printf( "%d\n", pos+2 );
    }
    return 0;
 }

D. 回文字D
尺取法加一个字符串hash，若是起始下标为i，长度为D的子串不为回文串，则统计个数直接加一，若是为回文串，则继续判起始下标为i+1，长度为D的子串是否为回文串，直到长度为D的子串不为回文串为止，统计个数加一。需要注意的是，若不为回文串，则将D-1个字符归为一串，下一次判的时候起始位置为这一次的终止位置即 i+D-1。

代码

#include <bits/stdc++.h>
#define ll long long
#define mem( f, x ) memset( f, x, sizeof(f))
#define pii pair<int,int>
#define fi first
#define se second
#define mk(x,y) make_pair(x,y)
#define pk push_back
typedef unsigned long long int ull;
using namespace std;
const int M = 1e7+5;
const int N = 1e7+5;
const int MOD = 1e9+7;
const int INF = 0x3f3f3f3f;
const int base=13331;  //玄学数字，这个数可以降低重复率
int m, n;
char s[N];
ull h1[N], h2[N], pw[N];

ull gethash1( int l, int r ){
    return h1[r]-h1[l-1]*pw[r-l+1];
}

ull gethash2( int l, int r ){
    return h2[r]-h2[l-1]*pw[r-l+1];
}

bool judge( int l, int r ){
    return gethash1(l, r) == gethash2(n-r+1, n-l+1);
}


int main( ){
    while( scanf( "%d %d", &n, &m ) != EOF ){
        scanf( "%s", s+1 );
        h1[0] = h2[0] = 0;
        pw[0] = 1;
        for( int i = 1; i <= n; i++ ){
            h1[i] = h1[i-1]*base+(s[i]-'a'+1);
            h2[i] = h2[i-1]*base+(s[n-i+1]-'a'+1);
            pw[i] = pw[i-1]*base;
        }
        int ans = 0;
        for( int st = 1, ed; st <= n; st = ed ){
            ed = st+m-1;
            while( ed <= n && judge( ed-m+1, ed ) ) ed++;
            ans++;
        }
        printf( "%d\n", ans );
    }
    return 0;
}