阶乘后面有k个0求满足的个数

这道题就不能像上面那样暴力做了，要二分

class Solution {
public:long long sum(int x){long long ans=0;//相当于有多少个是5的倍数，有多少个是25的倍数，有多少个是125的倍数....while(x){ans+=x/5;x/=5;}return ans;}long long cal(long long k){long long r=k*5;long long l=0;while(llong long mid=l+r>>1;if(sum(mid)>=k)r=mid;//找到最左的一个数字x，满足sum(x)大于等于kelse l=mid+1;}return r;}int preimageSizeFZF(int k) {return cal(k+1)-cal(k);}
};

E

记住当初的错误做法以及为什么错了

#include
#define int long long
using namespace std;
int mod=998244353;
int n,p,t,ans,A,B;
const int N=2e5+10;
int fac[N],inv[N];
int qpow(int a,int b){int ans=1;while(b){if(b&1)ans=ans*a%mod;b/=2;a=a*a%mod;}return ans;
}
int f[N];
int dfs(int x){if(x<=0)return 0;if(f[x])return f[x];return f[x]=(1+dfs(x-1)*A+dfs(x-2)*B)%mod;}
signed main(){ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);t=qpow(100,mod-2);cin>>n>>p;A=(100-p)*t%mod;B=p*t%mod;cout<

#include
using namespace std;
#define int long long
const int N=2e5+10;
int n,m,q,a,b,c,rt[N],cir[N],f[N];
typedef pairPII;
vectorg[N];
void dfs(int u,int fa){rt[u]=fa;for(auto [j,w]:g[u]){if(rt[j]==fa&&f[j]!=f[u]+w)cir[fa]=1;if(rt[j]==0)f[j]=f[u]+w,dfs(j,fa);}
}
signed main(){ios::sync_with_stdio(0);cin.tie(0);cin>>n>>m>>q;while(m--){cin>>a>>b>>c;g[a].push_back({b,c});g[b].push_back({a,-c});}for(int i=1;i<=n;i++){if(!rt[i])dfs(i,i);}while(q--){cin>>a>>b;if(rt[a]!=rt[b]){cout<<"nan"<<'\n';continue;}if(cir[rt[a]])cout<<"inf"<<'\n';else cout<