标签：style blog http io ar color os sp for
还是太弱啊 终测 C D都挂了 =_=
...porker写的C关于取模的运用 对于以后的题目 都有很好的 移植性 感觉主要是运用了 (a+b)%p = ( a%p+b%p)%p这个性质
贴下 3题代码 当时比赛的时候 可能有点难看 懒的改了
1 #include <cstdio> 2 #include <vector> 3 #include <algorithm> 4 using namespace std; 5 6 vector<int>ve[4]; 7 int a[10]; 8 9 int main() 10 { 11 int n , x , y , z , num; 12 while( ~scanf("%d",&n) ) 13 { 14 for( int i = 0 ; i<=3 ; i++ ) 15 ve[i].clear(); 16 for( int i = 1 ; i<=n ; i++ ) 17 { 18 scanf("%d",&num); 19 ve[ num ].push_back( i ); 20 } 21 x = ve[1].size(); 22 y = ve[2].size(); 23 z = ve[3].size(); 24 a[0] = x; 25 a[1] = y; 26 a[2] = z; 27 sort( a , a+3 ); 28 if( a[0]==0 ) 29 printf( "0\n" ); 30 else 31 { 32 printf( "%d\n",a[0] ); 33 for( int i = 0 ; i<a[0] ; i++ ) 34 { 35 printf( "%d %d %d\n",ve[1][i],ve[2][i],ve[3][i] ); 36 } 37 } 38 } 39 return 0; 40 }
B其实也蛮有意思的 先要 Hash找出队首 队尾元素 其实找一个就够了 他们出现的特点就是 只出现了1次 然后通过队首元素与第二个位置的元素的与其后继元素的Next关系
进行不断遍历 然后最后输出就可以了
1 #include <cstdio> 2 #include <cstring> 3 using namespace std; 4 5 const int size = 1000010; 6 int hash[size] , next[size] , pre[size]; 7 int ans[size]; 8 9 int main() 10 { 11 int n , x , y , cnt , first , last; 12 while( ~scanf("%d",&n) ) 13 { 14 memset( hash , 0 , sizeof(hash) ); 15 cnt = 0; 16 for( int i = 1 ; i<=n ; i++ ) 17 { 18 scanf( "%d %d",&x,&y ); 19 hash[x] ++; 20 hash[y] --; 21 next[x] = y; 22 pre[y] = x; 23 } 24 for( int i = 1 ; i<=size ; i++ ) 25 { 26 if( hash[i]==1 ) 27 { 28 ++ cnt; 29 first = i; 30 ans[1] = first; 31 } 32 else if( hash[i]==-1 ) 33 { 34 ++ cnt; 35 last = i; 36 ans[n] = last; 37 } 38 if( cnt==2 ) 39 break; 40 } 41 ans[2] = next[0]; 42 ans[n-1] = pre[0]; 43 int u = ans[1]; 44 int v = ans[2]; 45 int t = 3; 46 int k = 4; 47 bool flag = true; 48 while(1) 49 { 50 if( hash[ next[u] ]!=0 ) 51 { 52 flag = false; 53 } 54 else 55 { 56 ans[t] = next[u]; 57 t += 2; 58 u = next[u]; 59 } 60 if( hash[ next[v] ]!=0 ) 61 { 62 flag = false; 63 } 64 else 65 { 66 ans[k] = next[v]; 67 k += 2; 68 v = next[v]; 69 } 70 if( !flag ) 71 break; 72 } 73 for( int i = 1 ; i<=n-1 ; i++ ) 74 { 75 printf("%d ",ans[i] ); 76 } 77 printf("%d\n",ans[n]); 78 } 79 return 0; 80 }
1 #include <iostream> 2 #include <string> 3 using namespace std; 4 5 bool result[1000010]; 6 7 int main() { 8 string s; 9 bool flag; 10 cin >> s; 11 long long a, b; 12 cin >> a >> b; 13 long long temp = 1; 14 long long temp2 = 0; 15 flag = true; 16 for ( int i = s.size() - 1; i > 0; i--) { 17 temp %= b; 18 temp2 += temp * ( s[i] - ‘0‘ ); 19 temp2 %= b; 20 temp *= 10; 21 if ( temp2 == 0 && s[i] != ‘0‘ ) { 22 result[i] = true; 23 } 24 else { 25 result[i] = false; 26 } 27 } 28 temp2 = 0; 29 for (int i = 0; i < s.size()-1; i++) { 30 temp2 *= 10; 31 temp2 += s[i] - ‘0‘; 32 temp2 %= a; 33 if (temp2 == 0 && result[i + 1]) { 34 flag = false; 35 cout << "YES" << endl; 36 for (int j = 0; j <= i; j++) { 37 cout << s[j]; 38 } 39 cout << endl; 40 for (int j = i + 1; j < s.size(); j++) { 41 cout << s[j]; 42 } 43 cout << endl; 44 break; 45 } 46 } 47 if( flag ) 48 cout << "NO" << endl; 49 }
标签：style blog http io ar color os sp for
原文地址：http://www.cnblogs.com/radical/p/4118824.html
