2018ICPC 北京
A.0:27:15 solved by hl
感人的数据范围，每次连上边之后dfs一下前者看能不能找到自己就行
#include <iostream> #include <string> #include <map> #include <vector> #include <cstdio> using namespace std; int N; const int maxn = 110; map<string,int>Hash; vector<int>P[maxn]; bool vis[maxn]; int s; bool dfs(int t){ vis[t] = 1; for(int i = 0 ; i < P[t].size(); i ++){ int v = P[t][i]; if(v == s) return false; if(vis[v]) continue; if(!dfs(v)) return false; } return true; } int main() { while(~scanf("%d",&N)){ int cnt; cnt = 0; Hash.clear(); for(int i = 0 ; i < maxn; i ++) P[i].clear(); int flag = 0; for(int i = 1; i <= N; i ++){ string a,b; cin >> a >> b; if(flag) continue; if(!Hash[a]) Hash[a] = ++cnt; if(!Hash[b]) Hash[b] = ++cnt; P[Hash[a]].push_back(Hash[b]); for(int j = 0; j < maxn; j ++) vis[j] = 0; s = Hash[a]; if(!dfs(Hash[a])){ cout << a << " " << b << endl; flag = 1; } } if(!flag) puts("0"); } return 0; }
B.2:34:44(-6) solved by hl,gbs
debug题,两个人各写了一遍各WA了几发，最后提出0没有前导0的问题让两份代码都过了
先将所有串拼成一个很长的字符串，如果前一行末尾和后一行开头有一个不是数字，就要在两者之间加上空格
拼完之后就是一个简单的按题目统计了，对于每行出现的次数，要用一个数组记录一下每个字符串拼起来之后的断点
#include <iostream> #include <string> #include <map> #include <vector> #include <cstdio> #include <sstream> #include <algorithm> using namespace std; int N; const int maxn = 10010; string tmp,a; int id[maxn]; int num[maxn]; int main() { getline(cin,a); int cnt = 1; for(int i = 0 ; i < a.size(); i ++){ if(a[i] != ' ') id[1]++; } while(getline(cin,tmp)){ cnt++; if(!isdigit(a[a.size() - 1]) || !isdigit(tmp[0])){ a.push_back(' '); } id[cnt] = id[cnt - 1]; for(int i = 0 ; i < tmp.size(); i ++){ if(tmp[i] != ' ') id[cnt]++; } a = a + tmp; } // cout << a << endl; stringstream ss(a); int sum = 0; int fi = 0 ; while(ss >> tmp){ //cout << tmp << endl; int flag = 1; if(tmp[0] == '0' && tmp.size() > 1) flag = 0; if(islower(tmp[0]) || islower(tmp[tmp.size() - 1])) flag = 0; if(flag){ int p = upper_bound(id + 1,id + 1 + cnt,sum) - id; num[p]++; if(!fi) fi = 1; else cout << " "; cout << tmp; } sum += tmp.size(); } cout << endl; for(int i = 1; i <= cnt; i ++){ printf("%d\n",num[i]); } return 0; }
#include <iostream> #include <stdio.h> #include <string.h> using namespace std; char sum1[1000005]; int ans[3005]; int now_len1 = 0; int yuan; char lls1[1000005]; int now_row; int llen; bool if_first = true; inline bool if_d(char a) { if (a>'9') return false; if (a<'0') return false; return true; } bool if_OK; void end1() { lls1[llen] = 0; //cout<<lls1<<endl; if (llen != 0) { if (lls1[0] == '0' && llen != 1) { llen= 0; return ; } if (if_d(lls1[0])&& if_d(lls1[llen-1]) ) {//cout<<" && "<<llen<<endl; if (!if_first) { printf(" "); } if_first = false; printf("%s",lls1); ans[yuan]++; } } llen =0; } int main() { char a1; while(scanf("%c",&a1)!= EOF) { sum1[now_len1++] = a1; } //cout<<sum1; now_row = 0; llen =0; memset(ans,0,sizeof(ans)); for (int i=0;i<now_len1; i++) { //cout<<now_len1 if (sum1[i] == ' ') end1(); else if (sum1[i] == '\n') { if (sum1[i+1] != 0) now_row++; if (i == 0) end1(); else if(if_d(sum1[i-1]) && if_d(sum1[i+1])) { //cout<<"P"<<endl; } else end1(); } else { lls1[llen++] = sum1[i]; if (llen == 1) { yuan = now_row; if_OK = true; } if (!if_d(sum1[i])) { if_OK = false; } } } end1(); cout<<endl; for (int i=0; i<=now_row; i++) { cout<<ans[i]<<endl; } return 0; } /* a19 01 17b 12 bdc 13 23 14 0 bc */
D.1:30:06(-1) solved by zcz
是个构造,具体方法看代码
#include <iostream> #include <cstring> #include <cstdio> #include <vector> using namespace std; int vis[40]; vector<int> ans[2]; int main() { long long M; while(cin>>M) { if(M==0) { cout<<2<<endl; cout<<"1 1"<<endl; cout<<"2 2"<<endl; continue; } ans[0].clear(); ans[1].clear(); memset(vis,0,sizeof(vis)); int l=1; int cnt=0; while(M) { if(M&1) { cnt++; vis[l]=1; } l++; M>>=1; } int t=1; int i=0; int num=0; while(cnt) { if(vis[t]) { ans[0].push_back(i+1); ans[1].push_back(199); num++; //cout<<i+1<<' '<<199<<endl; vis[t]=0; cnt--; i=i+2; } else { ans[0].push_back(i+1); ans[1].push_back(i+2); num++; //cout<<i+1<<' '<<i+2<<endl; i=i+2; t++; } } ans[0].push_back(i); ans[1].push_back(i); num++; //cout<<i<<' '<<i<<endl; cout<<num<<endl; for(int j=0;j<num;j++) { cout<<ans[0][j]<<' '<<ans[1][j]<<endl; } } return 0; }
H.3:57:53 solved by hl
将所给模式串分成N + 1个模式串
其中一个是原来的，另一个是变了一个字母的
dp[i][j]表示i号模式串在以j位置结尾处出现的次数，每次容斥减掉前面出现过的次数
最后将每个dp[i][j] * 2 ^ (M - j)次方计入贡献，表示前面匹配成功之后后面就可以随便写
#include <iostream> #include <string> #include <map> #include <cstring> #include <vector> #include <cstdio> #include <sstream> #include <algorithm> #define LL long long using namespace std; int N,M; const int maxn = 110; char str[maxn][maxn]; LL dp[maxn][maxn]; LL pow2[maxn],sum[maxn]; int main() { int T; scanf("%d",&T); for(int i = 0; i <= 40; i ++) pow2[i] = (1LL << i); while(T--){ memset(dp,0,sizeof(dp)); memset(sum,0,sizeof(sum)); scanf("%d%d",&N,&M); scanf("%s",str[0] + 1); dp[0][N] = 1; for(int i = 1; i <= N; i ++){ for(int j = 1; j <= N ; j ++) str[i][j] = str[0][j]; if(str[i][i] == '0') str[i][i] = '1'; else str[i][i] = '0'; dp[i][N] = 1; } if(M < N){ puts("0"); continue; } sum[N] = (N + 1); for(int i = N + 1; i <= M ; i ++){ for(int j = 0 ; j <= N ; j ++){ dp[j][i] = pow2[i - N] - sum[i - N]; // cout << dp[j][i] << endl; for(int k = 0; k <= N ; k ++){ // if(k == j) continue; for(int l = 2; l <= N ; l ++){ int num = 0; for(int p = l; p <= N && num < 1; p ++){ if(str[k][p] != str[j][p - l + 1]) num++; } if(num < 1){ dp[j][i] -= dp[k][i - l + 1]; // cout << j << ' ' << k << " " << dp[k][i - l + 1] << " " << i - l + 1 << endl; } } } // cout << j << " " << i << " " << dp[j][i] << endl; sum[i] += dp[j][i]; } sum[i] += sum[i - 1] * 2; // cout << sum[i] << endl; } LL ans = 0; for(int i = 0; i <= N ; i ++){ for(int j = N ; j <= M ; j ++){ ans += dp[i][j] * pow2[M - j]; } } printf("%lld\n",ans); } return 0; }
I.1:19:19 solved by gbs
是个找规律，也可以推式子
具体规律看代码
#include <iostream> #include <stdio.h> #include <string.h> using namespace std; /* int anz[45]; int an0[45]; void dfs(int n) {\ int lln1 =n; INT T1; for (int i=0; i<20; i++) { if (lln1 <=anz[i]) { if (i == 0) cout<<lln1-1; else if (i == 1) cout<<lln1<<lln1; else { t1 =lln1/anz[i-2]; t1++; cout<<t1; ans[now_b++] = t1; dfs(); } return; } lln1 -= anz[i]; } }*/ char str1[100005]; int ans[200005]; int now_b; int main() { /*anz[0] = an0[0] = 10; anz[1] = 9; an0[1] = 10; for (int i=2; i<20; i++) { anz[i] = an0[i-2]*9; an0[i] = an0[i-2]*10; cout<<anz[i]<<endl; }*/ int T; //int n; cin >>T; while(T--) { scanf("%s",str1); int len1 = strlen(str1); int t1; if (len1 == 1) { t1 = str1[0]-'0'; t1--; cout<<t1<<endl; continue; } if (len1 == 2 &&str1[0] == '1'&&str1[1]=='0' ) { cout<<9<<endl; continue; } if (str1[0] == '1'&&str1[1]>='1' ) { t1 = str1[1] -'0'; printf("%d",t1); for (int i=2; i<len1; i++) printf("%c",str1[i]); for (int i=len1-1; i>=2; i--) printf("%c",str1[i]); printf("%d\n",t1); continue; } if (str1[0] == '1' ) { t1 = 9; printf("%d",t1); for (int i=2; i<len1; i++) printf("%c",str1[i]); for (int i=len1-2; i>=2; i--) printf("%c",str1[i]); printf("%d\n",t1); continue; } else { t1 = str1[0]-'1'; printf("%d",t1); for (int i=1; i<len1; i++) printf("%c",str1[i]); for (int i=len1-2; i>=1; i--) printf("%c",str1[i]); printf("%d\n",t1); continue; } } return 0; }
