1219 水~
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1 #include<stdio.h> 2 #include<string.h> 3 const int maxn = 30; 4 int a[ maxn ]; 5 char s[ 100005 ]; 6 int main(){ 7 while( gets( s )!=NULL ){ 8 memset( a,0,sizeof( a ) ); 9 int len=strlen( s ); 10 for( int i=0;i<len;i++ ){ 11 if( s[i]>='a'&&s[i]<='z' ) 12 a[ s[i]-'a' ]++; 13 } 14 for( int i=0;i<26;i++ ){ 15 printf("%c:%d\n",i+'a',a[i]); 16 } 17 printf("\n"); 18 } 19 return 0; 20 }
1222
我是找规律的。。。对于n，m，如果他们之间有公共约数，则说明在狼走第二圈的时候，会走重复的点。
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1 #include<stdio.h> 2 int gcd( int a,int b ){ 3 int r; 4 while( b ){ 5 r=a%b; 6 a=b; 7 b=r; 8 } 9 return a; 10 } 11 int main(){ 12 int t; 13 scanf("%d",&t); 14 while(t--){ 15 int n,m; 16 scanf("%d%d",&m,&n); 17 //while( m>n ) 18 //m-=n; 19 if( m==1||n==1 ){ 20 printf("NO\n"); 21 continue; 22 } 23 if( m==n ){ 24 printf("YES\n"); 25 continue; 26 } 27 if( gcd(n,m)!=1 ){ 28 printf("YES\n"); 29 } 30 else{ 31 printf("NO\n"); 32 } 33 } 34 return 0; 35 }
1224
SPFA 挺水的~~~
简单题吧
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1 #include<stdio.h> 2 #include<string.h> 3 #include<stdlib.h> 4 #include<algorithm> 5 #include<queue> 6 #include<stack> 7 using namespace std; 8 const int maxn = 105; 9 const int maxm = 10000; 10 const int inf = 99999999; 11 int cnt,head[ maxn ]; 12 struct node{ 13 int u,val,next; 14 }edge[ maxm ]; 15 int n,m; 16 int val[ maxn ]; 17 int dis[ maxn ],vis[ maxn ],path[ maxn ]; 18 19 void init(){ 20 cnt=0; 21 memset( head,-1,sizeof( head )); 22 } 23 24 void addedge( int a,int b,int c ){ 25 edge[ cnt ].u=b; 26 edge[ cnt ].val=c; 27 edge[ cnt ].next=head[ a ]; 28 head[ a ]=cnt++; 29 } 30 31 void bfs(){ 32 for( int i=1;i<=n+1;i++ ){ 33 vis[i]=0; 34 dis[i]=-inf; 35 path[i]=-1; 36 } 37 queue<int>q; 38 while( !q.empty() ) 39 q.pop(); 40 q.push( 1 ); 41 vis[1]=1; 42 dis[1]=0; 43 while( !q.empty() ){ 44 int now=q.front(); 45 q.pop(); 46 vis[ now ]=0; 47 for( int i=head[ now ];i!=-1;i=edge[ i ].next ){ 48 int next=edge[i].u; 49 if( dis[next]<dis[now]+val[next] ){ 50 dis[next]=dis[now]+val[next]; 51 path[next]=now; 52 if( vis[next]==0 ){ 53 vis[next]=1; 54 q.push(next); 55 } 56 } 57 } 58 } 59 } 60 void output(){ 61 printf("points : %d\n",dis[n+1]); 62 stack<int>s; 63 for( int i=n+1;i!=-1;i=path[i] ){ 64 s.push( i ); 65 } 66 printf("circuit : "); 67 printf("%d",s.top()); 68 s.pop(); 69 while( !s.empty() ){ 70 if( s.top()==(n+1) ) 71 printf("->%d",1); 72 else 73 printf("->%d",s.top()); 74 s.pop(); 75 } 76 printf("\n"); 77 } 78 int main(){ 79 int T; 80 scanf("%d",&T); 81 for( int ca=1;ca<=T;ca++ ){ 82 if( ca!=1 ) 83 printf("\n"); 84 scanf("%d",&n); 85 for( int i=1;i<=n;i++ ) 86 scanf("%d",&val[ i ]); 87 val[n+1]=0; 88 scanf("%d",&m); 89 int a,b; 90 init(); 91 while( m-- ){ 92 scanf("%d%d",&a,&b); 93 if( a<b ){ 94 addedge( a,b,0 ); 95 } 96 else if( a>b ){ 97 addedge( b,a,0 ); 98 } 99 } 100 bfs(); 101 printf("CASE %d#\n",ca); 102 //printf("%d\n",dis[n+1]); 103 output(); 104 } 105 return 0; 106 }
1225
字符串的水题~~~
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1 #include<stdio.h> 2 #include<string.h> 3 #include<algorithm> 4 using namespace std; 5 const int maxn = 5005; 6 struct node{ 7 char name[ 105 ]; 8 int win,lost,diff,score; 9 }team[ maxn ]; 10 int cnt; 11 void init( int n ){ 12 for( int i=0;i<=n;i++ ){ 13 team[ i ].win=team[ i ].lost=team[ i ].diff=team[ i ].score=0; 14 //memset( team[i].name,'\0',sizeof( team[i].name )); 15 } 16 cnt=0; 17 } 18 int find( char tt[] ){ 19 if( cnt==0 ){ 20 strcpy( team[0].name,tt ); 21 cnt++; 22 return 0; 23 } 24 int i; 25 for( i=0;i<cnt;i++ ){ 26 if( strcmp( tt,team[i].name )==0 ){ 27 return i; 28 } 29 } 30 if( i==cnt ){ 31 strcpy( team[cnt].name,tt ); 32 cnt++; 33 return cnt-1; 34 } 35 } 36 bool cmp( node a,node b ){ 37 if( a.score!=b.score ) 38 return a.score>b.score; 39 else if( a.diff!=b.diff ) 40 return a.diff>b.diff; 41 else if( a.win!=b.win ) 42 return a.win>b.win; 43 else if( strcmp( a.name,b.name )>0 ) 44 return false; 45 else return true; 46 } 47 int main(){ 48 int n; 49 while( scanf("%d",&n)!=EOF ){ 50 init( n ); 51 char n1[ 105 ],n2[ 105 ]; 52 int num1,num2; 53 for( int i=0;i<(n*(n-1));i++ ){ 54 scanf("%s VS %s %d:%d",n1,n2,&num1,&num2); 55 int f1=find( n1 ); 56 int f2=find( n2 ); 57 if( num1==num2 ){ 58 team[ f1 ].score++; 59 team[ f2 ].score++; 60 team[ f1 ].win+=num1; 61 team[ f1 ].lost+=num1; 62 team[ f2 ].win+=num2; 63 team[ f2 ].lost+=num2; 64 } 65 else if( num1>num2 ){ 66 team[ f1 ].win+=num1; 67 team[ f1 ].lost+=num2; 68 team[ f2 ].win+=num2; 69 team[ f2 ].lost+=num1; 70 team[ f1 ].score+=3; 71 } 72 else if( num1<num2 ){ 73 team[ f1 ].win+=num1; 74 team[ f1 ].lost+=num2; 75 team[ f2 ].win+=num2; 76 team[ f2 ].lost+=num1; 77 team[ f2 ].score+=3; 78 } 79 } 80 for( int i=0;i<n;i++ ){ 81 team[ i ].diff=team[ i ].win-team[ i ].lost; 82 } 83 sort( team,team+n,cmp ); 84 for( int i=0;i<n;i++ ){ 85 printf("%s %d\n",team[i].name,team[i].score); 86 } 87 printf("\n"); 88 } 89 return 0; 90 }
1226
BFS+5000个状态+mod！！！
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1 #include<stdio.h> 2 #include<stdlib.h> 3 #include<string> 4 #include<algorithm> 5 #include<math.h> 6 #include<iostream> 7 #include<queue> 8 using namespace std; 9 const int maxn = 24; 10 const int maxm = 5105; 11 const int inf = 99999999; 12 int n,m,c; 13 int a[ maxn ]; 14 int vis[ maxm ]; 15 int flag; 16 struct node{ 17 string ans; 18 int mod; 19 }; 20 string res; 21 queue<node>q; 22 void init2(){ 23 while( !q.empty() ) 24 q.pop(); 25 } 26 void init1(){ 27 memset( a,0,sizeof( a )); 28 memset( vis,0,sizeof( vis )); 29 flag=-1; 30 res.clear(); 31 } 32 void bfs(){ 33 init2(); 34 node now,next; 35 for( int i=1;i<16;i++ ){ 36 if( a[i] ){ 37 vis[ i%n ]=1; 38 if( i>=0&&i<=9 ){ 39 now.ans=""; 40 now.ans=(i+'0'); 41 now.mod=i%n; 42 q.push( now ); 43 } 44 else{ 45 now.ans=""; 46 now.ans=(i+'A'-10); 47 now.mod=i%n; 48 q.push( now ); 49 } 50 } 51 }//the init 52 while( !q.empty() ){ 53 now=q.front(),q.pop(); 54 if( flag==1&&now.ans.size()>res.size() ) 55 continue; 56 if( now.mod==0 ){ 57 if( flag==-1 ){ 58 flag=1; 59 res=now.ans; 60 } 61 else{ 62 if( now.ans.size()<res.size() ){ 63 res=now.ans; 64 } 65 else if( now.ans.size()==res.size()&&res>now.ans ){ 66 res=now.ans; 67 } 68 } 69 } 70 for( int i=0;i<16;i++ ){ 71 if( a[i] ){ 72 if( i<10 ){ 73 next=now; 74 next.ans+=( i+'0' ); 75 next.mod=(now.mod*c+i)%n; 76 if( ( next.ans.size()<=500&&vis[ next.mod ]==0 )||( next.mod==0 ) ){ 77 vis[ next.mod ]=1; 78 q.push( next ); 79 } 80 } 81 else{ 82 next=now; 83 next.ans+=(i+'A'-10); 84 next.mod=(now.mod*c+i)%n; 85 if( ( next.ans.size()<=500&&vis[ next.mod ]==0 )||( next.mod==0 ) ){ 86 vis[ next.mod ]=1; 87 q.push( next ); 88 } 89 } 90 } 91 } 92 } 93 } 94 95 int main(){ 96 int ca; 97 scanf("%d",&ca); 98 while( ca-- ){ 99 scanf("%d%d%d",&n,&c,&m); 100 init1(); 101 char ch[ 4 ]; 102 for( int i=0;i<m;i++ ){ 103 scanf("%s",ch); 104 if( ch[0]>='0'&&ch[0]<='9' ) 105 a[ ch[0]-'0' ]=1; 106 else 107 a[ ch[0]-'A'+10 ]=1; 108 } 109 if(n==0){ 110 // cout<<0<<endl; 111 if(a[0]==1) 112 cout<<0<<endl; 113 else 114 cout<<"give me the bomb please"<<endl; 115 continue; 116 } 117 bfs(); 118 if( flag==-1 ) 119 printf("give me the bomb please\n"); 120 else 121 cout<<res<<endl; 122 } 123 return 0; 124 }