Educational Codeforces Round 25
2017-07-17 19:26
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Educational Codeforces Round 25
A.Binary Protocol
题意分析
给出一种新的表示数字的方法,根据1和0来表示。首先是一个数字n,表示01串的长度。接下来是一个n个数的01串。表示的规则
1. 开头一定是1
2. 0表示分隔
3. 0分隔的区间内,1的个数就是这一位的数
4. 保证数字一定合法
代码总览
#include <bits/stdc++.h> #define nmax 100 using namespace std; char str[nmax]; int main() { int n; scanf("%d",&n); char t; getchar(); int num = 0; while(n--){ scanf("%c",&t); if(t == '1'){ num++; }else{ printf("%d",num); num = 0; } } printf("%d\n",num); return 0; }
B. Five-In-a-Row
题意分析
检查当前棋盘是否保证下一步X能获胜即可。感觉没有什么简单的方法,写了个大模拟。代码总览
#include <bits/stdc++.h> #define nmax 11 using namespace std; char mp[nmax][nmax]; bool iswin = false; char block[6]; bool check() { int numofX = 0 ,numofPoint = 0; for(int i= 0;i<=4;++i){ if(block[i] == 'X') numofX++; if(block[i] == '.') numofPoint++; } if(numofX == 4 && numofPoint == 1)return true; else return false; } bool check_row() { for(int i = 0;i<10;++i){ for(int j = 0;j<=5;++j){ for(int k = 0;k<=4;++k){ block[k] = mp[i][j+k]; } block[5] = '\0'; if(check()) return true; } } return false; } bool check_line() { for(int i = 0;i<10;++i){ for(int j = 0;j<=5;++j){ for(int k = 0;k<=4;++k){ block[k] = mp[j+k][i] ; } block[5] = '\0'; if(check()) return true; } } return false; } bool check_cross() { for(int i = 0; i<=5;++i){ for(int j = i;j<=5;++j){ for(int k = 0;k<5;++k){ block[k] = mp[j+k][j-i+k] ; } block[5] = '\0'; if( check()) return true; } } for(int i = 0; i<=5;++i){ for(int j = i;j<=5;++j){ for(int k = 0;k<5;++k){ block[k] = mp[j-i+k][j+k] ; } block[5] = '\0'; if(check()) return true; } } return false; } bool check_anticross() { for(int i = 0; i<=5;++i){ for(int j = i;j<=5;++j){ for(int k = 0;k<5;++k){ block[k] = mp[j+k][9-(j-i+k)] ; } block[5] = '\0'; if(check()) return true; } } for(int i = 0; i<=5;++i){ for(int j = i;j<=5;++j){ for(int k = 0;k<5;++k){ block[k] = mp[9-(j+k)][j-i+k]; } block[5] = '\0'; if(check()) return true; } } return false; } int main() { for(int i = 0;i<10;++i) scanf("%s",mp[i]); iswin = check_row(); if(!iswin) iswin = check_line(); if(!iswin) iswin = check_cross(); if(!iswin) iswin = check_anticross(); if(iswin == true) printf("YES\n"); else printf("NO\n"); return 0; }
C. Multi-judge Solving
题意分析
有个人想在AOJ上做题,每个题目有一个难度系数,现在这个人打算在AOJ上做n道题,这个人目前做出来的最高系数难度的题目是k,并且我们知道,对于难度系数为ai的题目,如果他已经做出来一道题d,且有2*d>=ai,他就能做出来ai这道题,否则的话,他就需要去BOJ上找一道题来做,使得他能做ai这道题。请问他至少要到BOJ上做几道题,才能全部做完n道题。先对难度系数排序,然后O(n)扫描,如果能做就做,并且更新目前最高的难度。如果不能做,就去BOJ找一道题做,并且更新难度系数。输出最后找的题目数量即可。
代码总览
#include <bits/stdc++.h> #define nmax 1005 using namespace std; int a[nmax]; int main() { int n,k; scanf("%d %d",&n,&k); for(int i = 1;i<=n;++i){ scanf("%d",&a[i]); } sort(a+1,a+n+1); int ans = 0; int nowmax = k; for(int i = 1;i<=n;++i){ if(a[i] <= nowmax*2){ if(a[i] >= nowmax){ nowmax = a[i]; } }else{ while(nowmax*2 < a[i]){ nowmax*=2; ans++; } if(a[i] >= nowmax){ nowmax = a[i]; } } } printf("%d\n",ans); return 0; }
D. Suitable Replacement
题意分析
有点乱代码总览
#include <bits/stdc++.h> #define fr first #define sc scanf #define pf printf #define se second #define ll long long #define pb push_back #define pr pair< int,int > #define fin(s) freopen( s, "r", stdin ) #define fout(s) freopen( s, "w", stdout ) #define TIME ios_base::sync_with_stdio(0) using namespace std; const ll INF = 1e9; const ll N = 1e3 + 1; const ll mod = 1e7 + 9; const long double eps = 1E-7; bool flag; int l[26]; string a, b; vector<int>V; int main() { TIME; cin >> a >> b; for( int i = 0; i < a.size(); i ++ ) { if( a[i] == '?' ) V.pb( i ); else l[a[i]-'a'] ++; } while( !flag ) { for( int i = 0; i < b.size(); i ++ ) { if( l[b[i]-'a'] ) l[b[i]-'a'] --; else { if( V.empty() ) flag = 1; else a[V.back()] = b[i], V.pop_back(); } } } cout << a << endl; }
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