Codeforces Beta Round #63 (Div. 2)
http://codeforces.com/contest/69
A
1 #include<bits/stdc++.h> 2 using namespace std; 3 #define lson l,mid,rt<<1 4 #define rson mid+1,r,rt<<1|1 5 #define sqr(x) ((x)*(x)) 6 #define pb push_back 7 #define eb emplace_back 8 #define maxn 1000006 9 #define rep(k,i,j) for(int k=i;k<j;k++) 10 typedef long long ll; 11 typedef unsigned long long ull; 12 13 int n; 14 struct sair{ 15 int x,y,z; 16 }a[105]; 17 18 int main(){ 19 #ifndef ONLINE_JUDGE 20 // freopen("input.txt","r",stdin); 21 #endif 22 std::ios::sync_with_stdio(false); 23 cin>>n; 24 for(int i=1;i<=n;i++){ 25 cin>>a[i].x>>a[i].y>>a[i].z; 26 } 27 int aa=0,bb=0,cc=0; 28 for(int i=1;i<=n;i++){ 29 aa+=a[i].x; 30 bb+=a[i].y; 31 cc+=a[i].z; 32 } 33 if(aa||bb||cc) cout<<"NO"<<endl; 34 else cout<<"YES"<<endl; 35 }
B
1 #include<bits/stdc++.h> 2 using namespace std; 3 #define lson l,mid,rt<<1 4 #define rson mid+1,r,rt<<1|1 5 #define sqr(x) ((x)*(x)) 6 #define pb push_back 7 #define eb emplace_back 8 #define maxn 1000006 9 #define rep(k,i,j) for(int k=i;k<j;k++) 10 typedef long long ll; 11 typedef unsigned long long ull; 12 13 int n,m; 14 int k[105],p[105]; 15 16 int main(){ 17 #ifndef ONLINE_JUDGE 18 // freopen("input.txt","r",stdin); 19 #endif 20 std::ios::sync_with_stdio(false); 21 int l,r,t,c; 22 int ans=0; 23 cin>>n>>m; 24 for(int i=0;i<m;i++){ 25 cin>>l>>r>>t>>c; 26 for(int j=l;j<=r;j++){ 27 if(t<k[j]||!k[j]){ 28 k[j]=t; 29 ans-=p[j]; 30 p[j]=c; 31 ans+=p[j]; 32 } 33 } 34 } 35 cout<<ans<<endl; 36 }
C
模拟
1 #include<bits/stdc++.h> 2 using namespace std; 3 #define lson l,mid,rt<<1 4 #define rson mid+1,r,rt<<1|1 5 #define sqr(x) ((x)*(x)) 6 #define pb push_back 7 #define eb emplace_back 8 #define maxn 1000006 9 #define rep(k,i,j) for(int k=i;k<j;k++) 10 typedef long long ll; 11 typedef unsigned long long ull; 12 13 char name1[110][110],name2[110][110]; 14 int with[110][110],have[110][110],n; 15 int g[110][110]; 16 char tname[110]; 17 char s[1000010]; 18 int Find(char *name) 19 { 20 for (int i=0;i<n;i++) 21 if (strcmp(name,name1[i])==0) return i; 22 return 0; 23 } 24 struct sair{ 25 char name[110]; 26 int num; 27 }; 28 sair p[110]; 29 bool cmp(sair x,sair y) 30 { 31 return strcmp(x.name,y.name)<0; 32 } 33 34 int main(){ 35 #ifndef ONLINE_JUDGE 36 // freopen("input.txt","r",stdin); 37 #endif 38 std::ios::sync_with_stdio(false); 39 int k,m,t; 40 scanf("%d%d%d%d",&k,&n,&m,&t); 41 for (int i=0;i<n;i++) 42 scanf("%s",name1[i]); 43 getchar(); 44 memset(g,0,sizeof(g)); 45 for (int ii=0;ii<m;ii++) 46 { 47 gets(s); 48 int len=strlen(s); 49 int cnt=0,i=0; 50 for (i=0;i<len&&s[i]!=':';i++) 51 name2[ii][cnt++]=s[i]; 52 name2[ii][cnt++]=0; 53 i+=2; 54 while (i<len) 55 { 56 cnt=0; 57 for (;s[i]!=' ';i++) 58 tname[cnt++]=s[i]; 59 tname[cnt]=0; 60 int id=Find(tname); 61 i++; 62 int num=0; 63 for (;s[i]!=0&&s[i]!=',';i++) 64 num=num*10+s[i]-'0'; 65 g[ii][id]=num; 66 i+=2; 67 } 68 } 69 memset(have,0,sizeof(have)); 70 memset(with,0,sizeof(with)); 71 for (int i=0;i<t;i++) 72 { 73 int tnum; 74 scanf("%d",&tnum); 75 scanf("%s",tname); 76 int id=Find(tname); 77 have[tnum][id]++; 78 for (int j=0;j<m;j++) 79 { 80 bool tfind=true; 81 for (int k=0;k<n;k++) 82 if (have[tnum][k]<g[j][k]) tfind=false; 83 if (tfind) 84 { 85 for (int k=0;k<n;k++) 86 have[tnum][k]-=g[j][k]; 87 with[tnum][j]++; 88 } 89 } 90 } 91 for (int i=1;i<=k;i++) 92 { 93 int cnt=0; 94 for (int j=0;j<n;j++) 95 if (have[i][j]!=0) 96 { 97 p[cnt].num=have[i][j]; 98 strcpy(p[cnt++].name,name1[j]); 99 } 100 for (int j=0;j<m;j++) 101 if (with[i][j]!=0) 102 { 103 p[cnt].num=with[i][j]; 104 strcpy(p[cnt++].name,name2[j]); 105 } 106 sort(p,p+cnt,cmp); 107 printf("%d ",cnt); 108 for (int j=0;j<cnt;j++) 109 printf("%s %d ",p[j].name,p[j].num); 110 } 111 }
D
记忆化搜索的博弈，和前天的E题很像
题意：给个初始点，然后两个人轮流移动一段距离，当点和原点的距离大于d时失败。题目中的点可以移动到y=x的对称点这条件没用，因为一个人使用了这个条件，另一个人也可以使用它使点回到刚刚的位置
1 #include<bits/stdc++.h> 2 using namespace std; 3 #define lson l,mid,rt<<1 4 #define rson mid+1,r,rt<<1|1 5 #define sqr(x) ((x)*(x)) 6 #define pb push_back 7 #define eb emplace_back 8 #define maxn 1000006 9 #define rep(k,i,j) for(int k=i;k<j;k++) 10 typedef long long ll; 11 typedef unsigned long long ull; 12 13 int d,n; 14 struct sair{ 15 int x,y; 16 }a[25]; 17 18 int book[405][405]; 19 20 bool dist(int a,int b){ 21 return (a-200)*(a-200)+(b-200)*(b-200) <= d*d; 22 } 23 24 int dfs(int x,int y){ 25 int xx,yy; 26 if(book[x][y]) return book[x][y]; 27 for(int i=1;i<=n;i++){ 28 xx=x+a[i].x; 29 yy=y+a[i].y; 30 if(dist(xx,yy)){ 31 if(2==dfs(xx,yy)){ 32 return book[x][y]=1; 33 } 34 } 35 } 36 return book[x][y]=2; 37 } 38 39 int main(){ 40 #ifndef ONLINE_JUDGE 41 // freopen("input.txt","r",stdin); 42 #endif 43 std::ios::sync_with_stdio(false); 44 int x,y; 45 cin>>x>>y>>n>>d; 46 for(int i=1;i<=n;i++){ 47 cin>>a[i].x>>a[i].y; 48 } 49 if(dfs(x+200,y+200)==1) cout<<"Anton"<<endl; 50 else cout<<"Dasha"<<endl; 51 52 }
E
题意:找出在一定区间内只出现过一次的最大的数
思路：建两个map，一个记录数的个数，另一个记录当前个数为1的值即可
1 #include<bits/stdc++.h> 2 using namespace std; 3 #define lson l,mid,rt<<1 4 #define rson mid+1,r,rt<<1|1 5 #define sqr(x) ((x)*(x)) 6 #define pb push_back 7 #define eb emplace_back 8 #define maxn 1000006 9 #define rep(k,i,j) for(int k=i;k<j;k++) 10 typedef long long ll; 11 typedef unsigned long long ull; 12 13 int n,k; 14 int a[100005]; 15 map<int,int>mp; 16 map<int,int>book; 17 map<int,int>::iterator it; 18 map<int,int>::reverse_iterator rit; 19 20 int main(){ 21 #ifndef ONLINE_JUDGE 22 // freopen("input.txt","r",stdin); 23 #endif 24 std::ios::sync_with_stdio(false); 25 cin>>n>>k; 26 for(int i=1;i<=n;i++) cin>>a[i]; 27 for(int i=1;i<=k;i++){ 28 mp[a[i]]++; 29 if(mp[a[i]]==2){ 30 book.erase(book.find(a[i])); 31 } 32 else if(mp[a[i]]==1){ 33 book[a[i]]=1; 34 } 35 } 36 if(book.size()==0){ 37 cout<<"Nothing"<<endl; 38 } 39 else{ 40 rit=book.rbegin(); 41 cout<<rit->first<<endl; 42 } 43 for(int i=k+1;i<=n;i++){ 44 mp[a[i-k]]--; 45 if(mp[a[i-k]]==0) book.erase(book.find(a[i-k])); 46 else if(mp[a[i-k]]==1) book[a[i-k]]=1; 47 mp[a[i]]++; 48 if(mp[a[i]]==1) book[a[i]]=1; 49 else if(mp[a[i]]==2)book.erase(book.find(a[i])); 50 if(book.size()==0){ 51 cout<<"Nothing"<<endl; 52 } 53 else{ 54 rit=book.rbegin(); 55 cout<<rit->first<<endl; 56 } 57 } 58 59 60 }