CartonsBase

Problem here Solution water #include <iostream> using namespace std ; int main(){ int n; while ( cin >> n){ while (n--){ int l; cin >> l; int t = (l/ 2 - 1 )*((l/ 2 - 1 )+ 1 ) / 2 ; if (t) cout << t << endl; else cout << 0 << endl; } } return 0 ; }

Problem here Solution *記得要開long long int #include <iostream> #include <memory.h> using namespace std ; long long int dp[ 51 ]; int main(){ int n; memset (dp, 0 , sizeof (dp)); dp[ 0 ] = 0 ; dp[ 1 ] = 1 ; dp[ 2 ] = 2 ; for ( int i = 3 ; i <= 50 ; i++) dp[i] = dp[i- 1 ] + dp[i- 2 ]; while ( cin >> n){ if (n == 0 ) break ; cout << dp[n] << endl; } return 0 ; }

Problem here Solution 輸入是從1開始數的，所以讀入時將u, v 都-1 先刷所有最短路 再找最小總和 #include <iostream> #include <string> #include <memory.h> #include <vector> #define max_num 9999 using namespace std ; int rat[ 30 ][ 30 ]; vector < string > names; int main(){ int n, m; int kase = 1 ; while ( cin >> n >> m){ if (n == 0 ) break ; for ( int i = 0 ; i < n; i++) for ( int j = 0 ; j < n; j++) if (i==j) rat[i][j] = rat[j][i] = 0 ; else rat[i][j] = rat[j][i] = max_num; for ( int i = 0 ; i < n; i++){ string input; cin >> input; names.push_back(input); } for ( int i = 0 ; i < m; i++){ int u, v, k; cin >> u >> v >> k; rat[u- 1 ][v- 1 ] = rat[v- 1 ][u- 1 ] = k; } for ( int i = ...

Problem  here Solution floyd刷所有點到其他點的最短路 再在上面找開始點到集合點的最大值 #include <iostream> #include <memory.h> #define MAX_NUM 1 << 29 using namespace std ; int main (){ int kase ; cin >> kase ; int cnt = 1 ; while ( kase --){ int n , r ; int rat [ 101 ][ 101 ]; cin >> n >> r ; for ( int i = 0 ; i < n ; i ++) for ( int j = 0 ; j < n ; j ++) if ( i == j ) rat [ i ][ j ] = rat [ j ][ i ] = 0 ; else rat [ i ][ j ] = MAX_NUM ; int s , e ; while ( r --){ int x , y ; cin >> x >> y ; rat [ x ][ y ] = rat [ y ][ x ] = 1 ; } for ( int i = 0 ; i < n ; i ++) for ( int j = 0 ; j < n ; j ++) for ( int k = 0 ; k < n ; k ++) ...

Problem  here Solution #include <iostream> #include <memory.h> #include <string> using namespace std ; int main (){ int n , m ; int len [ 1001 ]; while ( cin >> n >> m ){ if ( n == 0 && m == 0 ) break ; memset ( len , - 1 , sizeof ( len )); len [ n ] = 0 ; string output = "." ; while ( n ){ n *= 10 ; output += ( n / m ) + '0' ; n %= m ; if ( len [ n % m ] != - 1 ) break ; len [ n ] = output . length ()- 1 ; } for ( int i = 0 ; i < output . length (); i ++){ if ( i > 0 && ( i % 50 )== 0 ) cout << endl ; cout << output [ i ]; } cout << endl ; if ( n != 0 ) cout << "The last " ...

Problem  here Solution 自己的石子為點0，目標石子為1 生成MST直到點0點1連通 #include <iostream> #include <vector> #include <memory.h> #include <utility> #include <string> #include <math.h> #include <stdio.h> #include <algorithm> using namespace std ; vector < pair < float , pair < int , int > > > edges ; string spp ; float cal ( int x1 , int x2 , int y1 , int y2 ){ return ( float ) sqrt (( x1 - x2 )*( x1 - x2 ) + ( y1 - y2 )*( y1 - y2 )); } bool cmp ( pair < float , pair < int , int > > a , pair < float , pair < int , int > > b ){ return a . first < b . first ; } int pre [ 1000 ]; int find ( int x ){ int r = x ; while ( pre [ r ] != r ){ r = pre [ r ]; } int j = x , y ; while ( j != r ){ y = pre [ j ]; pre [ j ] = r ; j = y ; } return r ; } void ...