Codeforces Round #256 (Div. 2)

A - Rewards

#include<iostream>
#include<cstdio>
using namespace std;
int main(){
    int a[3], b[3], sum1 = 0, sum2 = 0;
    int n;
    for( int i = 0; i < 3; i++ ){
        cin>>a[i];
        sum1 += a[i];
    }
    for( int i = 0; i < 3; i++ ){
        cin>>b[i];
        sum2 += b[i];
    }
    int ans1, ans2;
    if( sum1 % 5 == 0 )  ans1 = sum1/5;
    else ans1 = sum1/5 + 1;
    if( sum2 % 10 == 0 ) ans2 = sum2/10;
    else ans2 = sum2/10 + 1;
    cin>>n;
    if( n >= (ans1 + ans2) ){
        printf( "YES
");
    }
    else printf( "NO
" );
    return 0;
}

B - Suffix Structures

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>

using namespace std;

int x[100], z[100];
char a[1000], b[1000];
int main(){
    cin >> a >> b;
    int n = strlen( a ), m = strlen( b );
    for( int i = 0; i < n; i++ ){
        x[ a[i]-'a' ]++;
    }
    for( int i = 0; i < n; i++ ){
        z[ b[i]-'a' ]++;
    }
    if( n == m ){
        bool flag = 0;
        for( int i = 0; i < 26; i++ ){
            if( x[i] != z[i] ) flag = 1;
        }
        if( !flag ){
            printf( "array
" ); return 0;
        }
        else {
            printf( "need tree
" ); return 0;
        }
    }
    if( n < m ){
        printf( "need tree
" ); return 0;
    }
    if( n > m ){
        bool flag = 0;
        for( int i = 0; i < 26; i++ ){
            if( z[i] > x[i] ) flag = 1;
        }
        if( flag ){
            printf( "need tree
" ); return 0;
        }
        if( !flag ){
            int i = 0, j = 0;
            while( j < m && i < n ){
                if( a[i] == b[j] ){
                    i++, j++;
                }
                else i++;
            }
            if( j == m ){
                printf( "automaton
" ); return 0;
            }
            else{
                printf( "both
" ); return 0;
            }
        }
    }
    return 0;
}

C - Painting Fence

刷木板。分治的思想，dfs(l, r, k), 看每一段是横着刷还是竖着刷比较忧。。

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdio>
#include<string>
#include<queue>
#include<cmath>
#include<vector>

using namespace std;

#define mnx 104000
#define ll long long
#define inf 0x3f3f3f3f
#define lson l, m, rt << 1
#define rson m+1, r, rt << 1 | 1

int a[mnx];
int dfs( int l, int r, int k ){
    if( l > r ) return 0;
    if( l == r ){
        return min( 1, a[l] - k );
    }
    int mn = min_element( a + l, a + r + 1 ) - a;
    return min( r - l + 1, a[mn] - k + dfs( l, mn - 1, a[mn] ) + dfs( mn + 1, r, a[mn] ) );
}
int main(){
    int n, mn = inf;
    scanf( "%d", &n );
    for( int i = 1; i <= n; i++ ){
        scanf( "%d", &a[i] );
        mn = min( mn, a[i] );
    }
    printf( "%d
", dfs( 1, n, 0 ) );
    return 0;
}

还有一种dp的做法，http://blog.csdn.net/u014733623/article/details/37927873 真巨神啊。。

说一下我的理解，dp[i][j]表示第i列以后的木板都刷完了（不包括第i列）且第j列是横着刷的，如果value[j]>=value[i]，则dp[i-1][j] = dp[i][i]，第i-1列刷完了，且第j列是横着刷的，因为第j列如果横着刷，肯定也能够把第i列刷了，所以会等于第i列以后刷完了且第i列是横着刷的。else， dp[i-1][j]=min(dp[i][j]+1,dp[i][i]+value[i]-value[j]),dp[i][j]+1,表示的是第i列以后的刷完了，且第j列是横着刷的，这时再把第i列竖着刷一次；dp[i][i]+value[i]-value[j]，表示的是第i列刷完了，且第i列是横着刷的(第i列实际上还没有刷),value[i]-value[j]这时如果刷第j列的话，第i列还要横着刷value[i] - value[j]次，所以 dp[i-1][j]=min(dp[i][j]+1,dp[i][i]+value[i]-value[j])

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdio>
#include<string>
#include<queue>
#include<cmath>
#include<vector>

using namespace std;

#define mnx 5050
#define ll long long
#define inf 0x3f3f3f3f
#define lson l, m, rt << 1
#define rson m+1, r, rt << 1 | 1

int a[mnx], dp[mnx][mnx];
int main(){
    int n;
    scanf( "%d", &n );
    for( int i = 1; i <= n; i++ ){
        scanf( "%d", &a[i] );
    }
    for( int i = n; i >= 1; i-- ){
        for( int j = 0; j < n; j++ ){
            if( a[j] >= a[i] ){
                dp[i-1][j] = dp[i][i];
            }
            else dp[i-1][j] = min( dp[i][j] + 1, dp[i][i] + a[i] - a[j] );
        }
    }
    printf( "%d
", dp[0][0] );
    return 0;
}

D - Multiplication Table

n*m的矩阵，每个格子的值是i*j，叫你找第k大的值。。二分的做法，对1，n*m二分，二分的值mid在第一行比它小的有 mid / 1个，第二行比它小的有mid / 2个，……，比它小的加起来如果比k小，就对 mid + 1， r 进行二分，否则就对 l, mid 进行二分

#include<iostream>
#include<cstring>
#include<string>
#include<cstdio>
#include<algorithm>
#include<cmath>

using namespace std;

#define inf 0x3f3f3f3f
#define eps 1e-8
#define maxn 111111
#define mod 1000007
#define ll long long

int main(){
    ll n, m, k;
    cin >> n >> m >> k;
    ll r = n * m, l = 1, mid;
    while( l < r ){
        mid = ( l + r ) >> 1;
        ll sum = 0;
        for( ll i = 1; i <= n; i++ ){
            sum += min( mid / i, m );
        }
        if( sum < k ){
            l = mid + 1;
        }
        else r = mid;
    }
    printf( "%I64d
", r );
    return 0;
}