2018ICPC 北京
作者：互联网
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;
}
A
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;
}
solved by hl
#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
*/
solved by gbs
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;
}
D
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;
}
H
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;
}
I
标签：cnt,int,北京,2018ICPC,t1,++,maxn,include 来源： https://www.cnblogs.com/Hugh-Locke/p/11396230.html