1 #include<iostream> 2 #include<cstdio> 3 using namespace std; 4 int main(){ 5 int a[3], b[3], sum1 = 0, sum2 = 0; 6 int n; 7 for( int i = 0; i < 3; i++ ){ 8 cin>>a[i]; 9 sum1 += a[i]; 10 } 11 for( int i = 0; i < 3; i++ ){ 12 cin>>b[i]; 13 sum2 += b[i]; 14 } 15 int ans1, ans2; 16 if( sum1 % 5 == 0 ) ans1 = sum1/5; 17 else ans1 = sum1/5 + 1; 18 if( sum2 % 10 == 0 ) ans2 = sum2/10; 19 else ans2 = sum2/10 + 1; 20 cin>>n; 21 if( n >= (ans1 + ans2) ){ 22 printf( "YES "); 23 } 24 else printf( "NO " ); 25 return 0; 26 }
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<string> 5 #include<algorithm> 6 7 using namespace std; 8 9 int x[100], z[100]; 10 char a[1000], b[1000]; 11 int main(){ 12 cin >> a >> b; 13 int n = strlen( a ), m = strlen( b ); 14 for( int i = 0; i < n; i++ ){ 15 x[ a[i]-'a' ]++; 16 } 17 for( int i = 0; i < n; i++ ){ 18 z[ b[i]-'a' ]++; 19 } 20 if( n == m ){ 21 bool flag = 0; 22 for( int i = 0; i < 26; i++ ){ 23 if( x[i] != z[i] ) flag = 1; 24 } 25 if( !flag ){ 26 printf( "array " ); return 0; 27 } 28 else { 29 printf( "need tree " ); return 0; 30 } 31 } 32 if( n < m ){ 33 printf( "need tree " ); return 0; 34 } 35 if( n > m ){ 36 bool flag = 0; 37 for( int i = 0; i < 26; i++ ){ 38 if( z[i] > x[i] ) flag = 1; 39 } 40 if( flag ){ 41 printf( "need tree " ); return 0; 42 } 43 if( !flag ){ 44 int i = 0, j = 0; 45 while( j < m && i < n ){ 46 if( a[i] == b[j] ){ 47 i++, j++; 48 } 49 else i++; 50 } 51 if( j == m ){ 52 printf( "automaton " ); return 0; 53 } 54 else{ 55 printf( "both " ); return 0; 56 } 57 } 58 } 59 return 0; 60 }
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。。
1 #include<iostream> 2 #include<cstring> 3 #include<algorithm> 4 #include<cstdio> 5 #include<string> 6 #include<queue> 7 #include<cmath> 8 #include<vector> 9 10 using namespace std; 11 12 #define mnx 104000 13 #define ll long long 14 #define inf 0x3f3f3f3f 15 #define lson l, m, rt << 1 16 #define rson m+1, r, rt << 1 | 1 17 18 int a[mnx]; 19 int dfs( int l, int r, int k ){ 20 if( l > r ) return 0; 21 if( l == r ){ 22 return min( 1, a[l] - k ); 23 } 24 int mn = min_element( a + l, a + r + 1 ) - a; 25 return min( r - l + 1, a[mn] - k + dfs( l, mn - 1, a[mn] ) + dfs( mn + 1, r, a[mn] ) ); 26 } 27 int main(){ 28 int n, mn = inf; 29 scanf( "%d", &n ); 30 for( int i = 1; i <= n; i++ ){ 31 scanf( "%d", &a[i] ); 32 mn = min( mn, a[i] ); 33 } 34 printf( "%d ", dfs( 1, n, 0 ) ); 35 return 0; 36 }
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])
1 #include<iostream> 2 #include<cstring> 3 #include<algorithm> 4 #include<cstdio> 5 #include<string> 6 #include<queue> 7 #include<cmath> 8 #include<vector> 9 10 using namespace std; 11 12 #define mnx 5050 13 #define ll long long 14 #define inf 0x3f3f3f3f 15 #define lson l, m, rt << 1 16 #define rson m+1, r, rt << 1 | 1 17 18 int a[mnx], dp[mnx][mnx]; 19 int main(){ 20 int n; 21 scanf( "%d", &n ); 22 for( int i = 1; i <= n; i++ ){ 23 scanf( "%d", &a[i] ); 24 } 25 for( int i = n; i >= 1; i-- ){ 26 for( int j = 0; j < n; j++ ){ 27 if( a[j] >= a[i] ){ 28 dp[i-1][j] = dp[i][i]; 29 } 30 else dp[i-1][j] = min( dp[i][j] + 1, dp[i][i] + a[i] - a[j] ); 31 } 32 } 33 printf( "%d ", dp[0][0] ); 34 return 0; 35 }
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
1 #include<iostream> 2 #include<cstring> 3 #include<string> 4 #include<cstdio> 5 #include<algorithm> 6 #include<cmath> 7 8 using namespace std; 9 10 #define inf 0x3f3f3f3f 11 #define eps 1e-8 12 #define maxn 111111 13 #define mod 1000007 14 #define ll long long 15 16 int main(){ 17 ll n, m, k; 18 cin >> n >> m >> k; 19 ll r = n * m, l = 1, mid; 20 while( l < r ){ 21 mid = ( l + r ) >> 1; 22 ll sum = 0; 23 for( ll i = 1; i <= n; i++ ){ 24 sum += min( mid / i, m ); 25 } 26 if( sum < k ){ 27 l = mid + 1; 28 } 29 else r = mid; 30 } 31 printf( "%I64d ", r ); 32 return 0; 33 }