1 void solve(){ 2 int n=read(); 3 string s; 4 cin>>s; 5 int ans=1; 6 for(int i=0;i<s.size()-1;i++){ 7 if(s[i]==s[i+1])ans=0; 8 } 9 puts(ans>0?"Yes":"No"); 10 }
void solve(){ int n=8,ansx,ansy; for(int i=1;i<=8;i++){ for(int j=1;j<=8;j++){ char a; cin>>a; if(a=='*'){ ansx=i; ansy=j; } } } char ans; ans=ansy-1+'a'; cout<<ans<<9-ansx<<endl; }
#define int long long map<int,int>mp; void solve(){ int n=read(),k=read(),ans=0; for(int i=1;i<=n;i++){ int x=read(); mp[x]++; if(mp[x+k]||mp[x-k]){ ans=1; } } puts(ans>0?"Yes":"No"); }
比赛的时候一直尝试增大m找符合的因数 然后tle
赛后看了题解才知道 可以遍历1-1e6的因数 找每个因数第一个大于等于m的值 记录最小的就可以得到答案
1 void solve(){ 2 int n=read(),m=read(); 3 int ans=INF; 4 for(int i=1;i<=min(n,1000000ll);i++){ 5 int j=m/i; 6 if(i*j<m){ 7 j++; 8 } 9 if(j<=n){ 10 ans=min(ans,i*j); 11 } 12 } 13 if(ans==INF){ 14 cout<<-1<<"\n"; 15 }else cout<<ans<<"\n"; 16 }
E - Transition Game
知道是求强连通 但是奈何本人太菜 只能赛后补知识点了/悲
int h[N], e[N], ne[N], idx, a[N]; int dfn[N], low[N], timestamp; int stk[N], top; bool in_stk[N]; int id[N], scc_cnt, siz[N]; void add(int a, int b){ e[idx] = b, ne[idx] = h[a], h[a] = idx ++; } void tarjan(int u){ dfn[u] = low[u] = ++ timestamp; stk[ ++ top] = u, in_stk[u] = true; for (int i = h[u]; ~i; i = ne[i]){ int j = e[i]; if (!dfn[j]){ tarjan(j); low[u] = min(low[u], low[j]); } else if (in_stk[j]) low[u] = min(low[u], dfn[j]); } if (dfn[u] == low[u]){ ++ scc_cnt; int y; do { y = stk[top -- ]; in_stk[y] = false; id[y] = scc_cnt; siz[scc_cnt]++; } while (y != u); } } void solve(){ idx = 0; memset(h, -1, sizeof h); int n=read(),ans=0; for(int i=1;i<=n;i++){ a[i]=read(); if(i!=a[i]){ add(i,a[i]); } } for(int i=1;i<=n;i++){ if(a[i]!=i&&!dfn[i]){ tarjan(i); } } for(int i=1;i<=n;i++){ if(a[i]==i||siz[id[i]]>1){ ans++; } } cout<<ans<<endl; }
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