省選模擬18
阿新 • • 發佈:2022-02-19
改題改到了很晚很晚,題解就不寫的那麼詳細了
總之考場上是渾渾噩噩,於是只有暴力分
T1 進位制轉換
說白了就是一個卡範圍列舉......
AC_code
#include<bits/stdc++.h> using namespace std; #define int long long #define fo(i,x,y) for(int i=(x);i<=(y);i++) #define fu(i,x,y) for(int i=(x);i>=(y);i--) int read(){ int s=0,t=1;char ch=getchar(); while(!isdigit(ch)){if(ch=='-')t=-1;ch=getchar();} while(isdigit(ch)){s=s*10+ch-'0';ch=getchar();} return s*t; } const int N=30; int T,n,m,b; int sn[N],sm[N]; bool ck(int x){int now=n; for(sn[0]=1;now;sn[0]++,now/=x){ sn[sn[0]]=now%x; if(sn[sn[0]]>9)return false; }sn[0]--; if(sn[0]<sm[0])return false; if(sn[0]>sm[0])return true; fu(i,sn[0],1){ if(sn[i]<sm[i])return false; else if(sn[i]>sm[i])return true; } return true; } void work(){ n=read();m=read();int mm=m; for(sm[0]=1;mm;sm[0]++,mm/=10)sm[sm[0]]=mm%10;sm[0]--; for(int c=log10(m);c<9;c++)for(int a=1;a<=9;a++){ for(int br=powl(n/a,1.0L/c),bl=max((int)powl((n-(powl(br,c)-1)/(br-1)*9)/a,1.0L/c),(int)10);bl<=br;br--) if(ck(br)){printf("%lld\n",br);return ;} }b=101; while(--b)if(ck(b)){printf("%lld\n",b);return ;} } signed main(){ freopen("number.in","r",stdin); freopen("number.out","w",stdout); T=read(); while(T--)work(); return 0; }
T2 遇到困難睡大覺
這個不會
T3 張士超你昨天晚上到底把我家鑰匙放在哪了
發現可以列舉有幾個集合超過了限制然後容斥
其實是兩個容斥的合併,但是從合併的角度來看並不好理解,直接容斥就好了
AC_code
#include<bits/stdc++.h> using namespace std; #define int long long #define fo(i,x,y) for(int i=(x);i<=(y);i++) #define fu(i,x,y) for(int i=(x);i>=(y);i--) int read(){ int s=0,t=1;char ch=getchar(); while(!isdigit(ch)){if(ch=='-')t=-1;ch=getchar();} while(isdigit(ch)){s=s*10+ch-'0';ch=getchar();} return s*t; } const int N=205; const int mod=998244353; int ksm(int x,int y){ int ret=1; while(y){ if(y&1)ret=ret*x%mod; x=x*x%mod;y>>=1; }return ret; } int n,m,al,d,a[N],p[N]; int f[N][N][N],ans; int inv[N],c1[N],c2[N]; signed main(){ freopen("key.in","r",stdin); freopen("key.out","w",stdout); m=read();d=read();al=read();n=read(); fo(i,1,m)a[i]=(read()+1)/d,p[i]=read(),inv[i]=ksm(i,mod-2); f[0][0][0]=1; fo(i,1,m)fu(j,al/d,0)fu(k,al/d,0)fu(m1,m,0){ if(k*d>al-n||k*d+j*d>al)continue; // cout<<i<<" "<<j<<" "<<k<<" "<<m1<<" "<<f[j][k][m1]<<endl; f[j+a[i]][k][m1+1]=(f[j+a[i]][k][m1+1]-f[j][k][m1]*p[i]%mod+mod)%mod; f[j][k][m1+1]=(f[j][k][m1+1]+f[j][k][m1]*p[i])%mod; f[j][k+a[i]][m1]=(f[j][k+a[i]][m1]-f[j][k][m1]*(1-p[i]+mod)%mod+mod)%mod; f[j][k][m1]=f[j][k][m1]*(1-p[i]+mod)%mod; } fo(j,0,al/d)fo(k,0,al/d)fo(m1,1,m){ if(k*d>al-n||k*d+j*d>al)continue; int n1=max(n-j*d,0ll),n2=al-n1-(j+k)*d; if(!n1){int res=1; fo(i,1,m-1)res=res*(i+n2)%mod*inv[i]%mod; ans=(ans+f[j][k][m1]*res)%mod;continue; } fo(i,1,m1)c1[i]=c2[i]=0; c1[1]=1;fo(i,2,m1)c1[i]=c1[i-1]*(n1-1+i-1)%mod*inv[i-1]%mod; c2[m1]=1;fo(i,n2+1,n2+m-m1)c2[m1]=c2[m1]*i%mod*inv[i-n2]%mod; fu(i,m1-1,1)c2[i]=c2[i+1]*(n2+m-i)%mod*inv[m-i]%mod; fo(i,1,m1)ans=(ans+f[j][k][m1]*c1[i]%mod*c2[i])%mod; } printf("%lld",ans); }