ICPC North America Qualifier Contest 2015 按通过率从 J开始

J. Torn To Pieces

思路很简单，主要是怎么实现：

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define ls (o<<1)
#define rs (o<<1|1)
#define pb push_back
const double PI= acos(-1.0);
const int M = 100+7;
/*
int head[M],cnt;
void init(){cnt=0,memset(head,-1,sizeof(head));}
struct EDGE{int to,nxt,val;}ee[M*2];
void add(int x,int y){ee[++cnt].nxt=head[x],ee[cnt].to=y,head[x]=cnt;}
*/
vector<int>G[M];
map<string,map<string,int> >mp;
map<string,int>id;
map<int,string>di;
vector<int>p;
string st,ed;
int vs[M];
bool f;
void dfs(int u,int fa)
{vs[u]=1;if(u==id[ed]){if(p.size()==0)return ;for(int i=0;i<p.size();i++)cout<<di[p[i]]<<" ";cout<<endl;f=true;return ;}for(auto v:G[u]){if(v==fa||vs[v])continue;p.pb(v);dfs(v,u);p.pop_back();}
}
char S[1100];
int main()
{int n,sz=0;// freopen("test_15.in","r",stdin);cin>>n;getchar();for(int i=1;i<=n;i++){int nm=1;cin.getline(S,1000);int l=strlen(S);string s[40];for(int j=0;j<l;j++){if(S[j]==' '){if(id.find(s[nm])==id.end())id[s[nm]]=++sz;di[id[s[nm]]]=s[nm];++nm; }else s[nm]+=S[j];}if(id.find(s[nm])==id.end())id[s[nm]]=++sz;di[id[s[nm]]]=s[nm];for(int j=2;j<=nm;j++){int u=id[s[j]];int v=id[s[1]];G[u].pb(v);G[v].pb(u);}}cin>>st>>ed;p.pb(id[st]);if(id[st])dfs(id[st],-1);if(!f)cout<<"no route found"<<endl;return 0;
}

A. All about that base

模拟+暴力就完事了注意特判

#include <cstdio>
#include<cstring>
using namespace std;
typedef unsigned long long ll;
const int M = 110+7;
char s[M];
char pr[40]={"0123456789abcdefghijklmnopqrstuvwxyz0"};
ll a[3][40];
int gt(char ch)
{if(ch<='9'&&ch>='0')return ch-'0';return ch-'a'+10;
}
int main()
{int t;scanf("%d",&t);ll up=1;for(int i=1;i<=32;i++)up*=2;up--;while(t--){scanf(" %[^\n]",s);int l=strlen(s);int tp=0;char op;memset(a,0,sizeof(a));for(int i=0;i<l;i++){if(s[i]==' '){if(tp==0)op=s[i+1];tp++;i+=2;continue;}for(int j=1;j<=36;j++){if(a[tp][j]==-1)continue;a[tp][j]*=j;ll z=gt(s[i]);a[tp][j]+=z;if(a[tp][j]>up||z>j||z==j&&j!=1||z==0&&j==1)a[tp][j]=-1;}}int nm=0;for(int i=1;i<=36;i++){bool f=true;for(int j=0;j<=2;j++){if(a[j][i]==-1)f=false;}//   printf("%d  %lld %lld %lld\n",i,a[0][i],a[1][i],a[2][i]);if(!f)continue;ll w;if(op=='+'&&a[0][i]+a[1][i]==a[2][i]);else if(op=='-'&&a[0][i]-a[1][i]==a[2][i]);else if(op=='*'&&a[0][i]*a[1][i]==a[2][i]);else if(op=='/'&&a[2][i]*a[1][i]==a[0][i]);else f=false;if(f)printf("%c",pr[i]),nm++;}if(nm==0)printf("invalid");puts("");}return 0;
}

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