CodeForces1165

CodeForces1165A

CodeForces1165A
水题,数一数后\(x\)位里的\(1\),注意\(y+1\)位是不是\(1\)就行了.

#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <string>
#include <vector>
#include <queue>
#include <cmath>
#include <ctime>
#include <map>
#include <set>
#define MEM(x,y) memset ( x , y , sizeof ( x ) )
#define rep(i,a,b) for (int i = (a) ; i <= (b) ; ++ i)
#define per(i,a,b) for (int i = (a) ; i >= (b) ; -- i)
#define pii pair < int , int >
#define X first
#define Y second
#define rint read<int>
#define int long long
#define pb push_back

using std::queue ;
using std::set ;
using std::pair ;
using std::max ;
using std::min ;
using std::priority_queue ;
using std::vector ;
using std::swap ;
using std::sort ;
using std::unique ;
using std::greater ;

template < class T >
    inline T read () {
        T x = 0 , f = 1 ; char ch = getchar () ;
        while ( ch < '0' || ch > '9' ) {
            if ( ch == '-' ) f = - 1 ;
            ch = getchar () ;
        }
       while ( ch >= '0' && ch <= '9' ) {
            x = ( x << 3 ) + ( x << 1 ) + ( ch - 48 ) ;
            ch = getchar () ;
       }
   return f * x ;
}

const int N = 2e5 + 100 ;

int n , x , y , ans ;
bool v[N] ; char s[N] ;

signed main (int argc , char * argv[]) {
    n = rint () ; x = rint () ; y = rint () ;
    scanf ("%s" , s + 1 ) ; rep ( i , 1 , n ) v[i] = s[i] == '1' ;
    per ( i , n , n - x + 1 ) if ( v[i] ) ++ ans ;
    if ( ! v[n-y] ) ++ ans ; else -- ans ;
    printf ("%lld\n" , ans ) ;
    system ("pause") ; return 0 ;
}

CodeForces1165B

水题,把比赛排个序,能打就打.

#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <string>
#include <vector>
#include <queue>
#include <cmath>
#include <ctime>
#include <map>
#include <set>
#define MEM(x,y) memset ( x , y , sizeof ( x ) )
#define rep(i,a,b) for (int i = (a) ; i <= (b) ; ++ i)
#define per(i,a,b) for (int i = (a) ; i >= (b) ; -- i)
#define pii pair < int , int >
#define X first
#define Y second
#define rint read<int>
#define int long long
#define pb push_back

using std::queue ;
using std::set ;
using std::pair ;
using std::max ;
using std::min ;
using std::priority_queue ;
using std::vector ;
using std::swap ;
using std::sort ;
using std::unique ;
using std::greater ;

template < class T >
    inline T read () {
        T x = 0 , f = 1 ; char ch = getchar () ;
        while ( ch < '0' || ch > '9' ) {
            if ( ch == '-' ) f = - 1 ;
            ch = getchar () ;
        }
       while ( ch >= '0' && ch <= '9' ) {
            x = ( x << 3 ) + ( x << 1 ) + ( ch - 48 ) ;
            ch = getchar () ;
       }
   return f * x ;
}

const int N = 2e6 + 100 ;

int n , v[N] , ans ;

signed main (int argc , char * argv[]) {
    n = rint () ; rep ( i , 1 , n ) v[i] = rint () ;
    sort ( v + 1 , v + n + 1 ) ; int now = 1 , cur = 1 ;
    while ( true ) {
        if ( v[cur] >= now ) ++ ans ;
        else {
            while ( v[cur] < now && cur <= n ) ++ cur ;
            if ( cur > n ) break ; ++ ans ;
        }
        ++ now ; ++ cur ;
    }
    printf ("%lld\n" , ans ) ;
    system ("pause") ; return 0 ;
}

CodeForces1165C

水题,从前向后扫原串,能取就取,最后注意取出来的长度就行了.

#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <string>
#include <vector>
#include <queue>
#include <cmath>
#include <ctime>
#include <map>
#include <set>
#define MEM(x,y) memset ( x , y , sizeof ( x ) )
#define rep(i,a,b) for (int i = (a) ; i <= (b) ; ++ i)
#define per(i,a,b) for (int i = (a) ; i >= (b) ; -- i)
#define pii pair < int , int >
#define X first
#define Y second
#define rint read<int>
#define int long long
#define pb push_back

using std::queue ;
using std::set ;
using std::pair ;
using std::max ;
using std::min ;
using std::priority_queue ;
using std::vector ;
using std::swap ;
using std::sort ;
using std::unique ;
using std::greater ;

template < class T >
    inline T read () {
        T x = 0 , f = 1 ; char ch = getchar () ;
        while ( ch < '0' || ch > '9' ) {
            if ( ch == '-' ) f = - 1 ;
            ch = getchar () ;
        }
       while ( ch >= '0' && ch <= '9' ) {
            x = ( x << 3 ) + ( x << 1 ) + ( ch - 48 ) ;
            ch = getchar () ;
       }
   return f * x ;
}

const int N = 2e6 + 100 ;

int n , ans ;
char s[N] , t[N] ;

signed main (int argc , char * argv[]) {
    n = rint () ; scanf ("%s" , s + 1 ) ;
    if ( ! ( n & 1 ) ) {
        bool f = true ;
        for (int i = 1 ; i <= n - 1 ; i += 2)
            if ( s[i] == s[i+1] ) { f = false ; break ; }
        if ( f ) {
            puts ("0") ;
            rep ( i , 1 , n ) putchar ( s[i] ) ; putchar ( 10 ) ;
            system ("pause") ; return 0 ;
        }
    }
    int cur = 0 ;
    rep ( i , 1 , n ) {
        if ( cur & 1 ) {
            while ( s[i] == t[cur] && i <= n ) ++ i ;
            if ( i > n ) { -- cur ; break ; }
            t[++cur] = s[i] ;
        } else t[++cur] = s[i] ;
    }
    if ( cur & 1 ) -- cur ;
    printf ("%lld\n" , n - cur ) ;
    if ( cur ) rep ( i , 1 , cur ) putchar ( t[i] ) ; putchar(10) ;
    system ("pause") ; return 0 ;
}

CodeForces1165D

水题,把给定的因子排个序,取\(d_1\times d_n\)为假定答案,然后从两侧向中间扫,遇到矛盾直接\(-1\).
如果通过了上面的检测,就再\(\Theta(\sqrt{n})\)枚举因子,判断是否全部出现即可.

#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <string>
#include <vector>
#include <queue>
#include <cmath>
#include <ctime>
#include <map>
#include <set>
#define MEM(x,y) memset ( x , y , sizeof ( x ) )
#define rep(i,a,b) for (int i = (a) ; i <= (b) ; ++ i)
#define per(i,a,b) for (int i = (a) ; i >= (b) ; -- i)
#define pii pair < int , int >
#define X first
#define Y second
#define rint read<int>
#define int long long
#define pb push_back

using std::queue ;
using std::set ;
using std::pair ;
using std::max ;
using std::min ;
using std::priority_queue ;
using std::vector ;
using std::swap ;
using std::sort ;
using std::unique ;
using std::greater ;

template < class T >
    inline T read () {
        T x = 0 , f = 1 ; char ch = getchar () ;
        while ( ch < '0' || ch > '9' ) {
            if ( ch == '-' ) f = - 1 ;
            ch = getchar () ;
        }
       while ( ch >= '0' && ch <= '9' ) {
            x = ( x << 3 ) + ( x << 1 ) + ( ch - 48 ) ;
            ch = getchar () ;
       }
   return f * x ;
}

const int N =  521 ;
const int M = 1e7 + 100 ;
int n , d[N] , ans ;
bool mk[M] ;

signed main (int argc , char * argv[]) {
    int T = rint () ;
    while ( T -- ) {
        n = rint () ; MEM ( mk , 0 ) ;
        rep ( i , 1 , n ) d[i] = rint () , mk[d[i]] = true ;
        sort ( d + 1 , d + n + 1 ) ; bool f = true ;
        int cmp = d[1] * d[n] ; int maxf = n >> 1 ;
        for (int i = 2 ; i <= maxf ; ++ i)
            if ( d[i] * d[n-i+1] != cmp ) { f = false ; break ; }
        if ( n & 1 ) if ( d[maxf+1] * d[maxf+1] != cmp ) f = false ;
        if ( f ) for (int i = 2 ; i * i <= cmp + 1 ; ++ i)
            if ( cmp % i == 0 && ( ! mk[i] || ! mk[cmp/i] ) ) f = false ;
        if ( ! f ) puts ("-1") ; else printf ("%lld\n" , cmp ) ;
    }
    system ("pause") ; return 0 ;
}

CodeForces1165E

有点东西的题目.
考虑每个位置的贡献,你发现每个位置都是独立的,然后分别考虑贡献就行了.
贡献就是经过它的区间个数乘上\(a\)数组,然后把得到的贡献数组和\(b\)反向排序,对应位置统计即可.

#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <string>
#include <vector>
#include <queue>
#include <cmath>
#include <ctime>
#include <map>
#include <set>
#define MEM(x,y) memset ( x , y , sizeof ( x ) )
#define rep(i,a,b) for (int i = (a) ; i <= (b) ; ++ i)
#define per(i,a,b) for (int i = (a) ; i >= (b) ; -- i)
#define pii pair < int , int >
#define X first
#define Y second
#define rint read<int>
#define int long long
#define pb push_back

using std::queue ;
using std::set ;
using std::pair ;
using std::max ;
using std::min ;
using std::priority_queue ;
using std::vector ;
using std::swap ;
using std::sort ;
using std::unique ;
using std::greater ;

template < class T >
    inline T read () {
        T x = 0 , f = 1 ; char ch = getchar () ;
        while ( ch < '0' || ch > '9' ) {
            if ( ch == '-' ) f = - 1 ;
            ch = getchar () ;
        }
       while ( ch >= '0' && ch <= '9' ) {
            x = ( x << 3 ) + ( x << 1 ) + ( ch - 48 ) ;
            ch = getchar () ;
       }
   return f * x ;
}

const int N = 2e5 + 100 ;
const int mod = 998244353 ;

int n , a[N] , b[N] , ans , pos[N] ;

inline bool cmp (int x , int y) { return x > y ; }

signed main (int argc , char * argv[]) {
    n = rint () ;
    rep ( i , 1 , n ) a[i] = rint () ; rep ( i , 1 , n ) b[i] = rint () ;
    rep ( i , 1 , n ) pos[i] = a[i] * ( ( i - 1 ) * ( n - i ) + n ) ;
    sort ( pos + 1 , pos + n + 1 , cmp ) ; sort ( b + 1 , b + n + 1 ) ;
    rep ( i , 1 , n ) ans = ( ans + pos[i] % mod * b[i] % mod ) % mod ;
    printf ("%lld\n" , ans ) ; system ("pause") ; return 0 ;
}