A:签到。
#include<bits/stdc++.h> using namespace std; #define ll long long #define inf 1000000010 char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<'0'||c>'9')) c=getchar();return c;} int gcd(int n,int m){return m==0?n:gcd(m,n%m);} int read() { int x=0,f=1;char c=getchar(); while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();} while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar(); return x*f; } int T,a[6],cnt[6]; signed main() { T=read(); while (T--) { for (int i=1;i<=5;i++) a[i]=read(); memset(cnt,0,sizeof(cnt)); for (int i=1;i<=5;i++) cnt[a[i]]++; int mx=0; for (int i=1;i<=5;i++) if (cnt[i]>cnt[mx]) mx=i; cout<<mx<<endl; } return 0; //NOTICE LONG LONG!!!!! }
B:k是奇数时函数值均为1,k是偶数时每k+1个出现一个0。
#include<bits/stdc++.h> using namespace std; #define ll long long #define inf 1000000010 char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<'0'||c>'9')) c=getchar();return c;} int gcd(int n,int m){return m==0?n:gcd(m,n%m);} ll read() { ll x=0,f=1;char c=getchar(); while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();} while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar(); return x*f; } int T; ll l,r,k; ll f(ll x) { return x-(x+1)/(k+1); } signed main() { T=read(); while (T--) { l=read(),r=read(),k=read(); if (k&1) cout<<r-l+1<<endl; else cout<<f(r)-f(l-1)<<endl; } return 0; //NOTICE LONG LONG!!!!! }
C:暴力枚举横竖各切多少刀,将矩阵压成一行可以得到该情况下列的划分位置,压成一列可以得到该情况下行的划分位置,然后二维前缀和暴力验证即可。
#include<bits/stdc++.h> using namespace std; #define ll long long #define inf 1000000010 #define N 1010 char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<'0'||c>'9')) c=getchar();return c;} int gcd(int n,int m){return m==0?n:gcd(m,n%m);} ll read() { ll x=0,f=1;char c=getchar(); while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();} while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar(); return x*f; } int T,n,m,k,tot,a[N][N],s[N],s2[N],s3[N][N],ans[N<<1],way[N<<1]; int calc(int x,int y,int l,int r) { return s3[y][r]-s3[x-1][r]-s3[y][l-1]+s3[x-1][l-1]; } signed main() { T=read(); while (T--) { n=read(),m=read(),k=read();tot=0; for (int i=1;i<=n;i++) for (int j=1;j<=m;j++) { char c=getc(); if (c=='0') a[i][j]=0;else a[i][j]=1,tot++; } for (int i=1;i<=n;i++) { s[i]=s[i-1]; for (int j=1;j<=m;j++) if (a[i][j]) s[i]++; } for (int i=1;i<=m;i++) { s2[i]=s2[i-1]; for (int j=1;j<=n;j++) if (a[j][i]) s2[i]++; } for (int i=1;i<=n;i++) for (int j=1;j<=m;j++) s3[i][j]=s3[i-1][j]+s3[i][j-1]-s3[i-1][j-1]+1-a[i][j]; tot=n*m-tot; bool isac=0; for (int i=1;i<=k;i++) ans[i]=2010; for (int i=0;i<=k;i++) if (tot%((i+1)*(k-i+1))==0) { int sum=tot/((i+1)*(k-i+1)); int last=0; int cnt=0; bool flag=1; for (int j=1;j<n;j++) if (cnt<i) { if ((j-last)*m-(s[j]-s[last])==sum*(k-i+1)) { way[++cnt]=j; last=j; } else if ((j-last)*m-(s[j]-s[last])>sum*(k-i+1)) {flag=0;break;} } if (cnt<i) flag=0; if (flag) { last=0; for (int j=1;j<m;j++) if (cnt<k) { if ((j-last)*n-(s2[j]-s2[last])==sum*(i+1)) { way[++cnt]=n+j-1; last=j; } else if ((j-last)*n-(s2[j]-s2[last])>sum*(i+1)) {flag=0;break;} } } if (cnt<k) flag=0; if (flag) { bool u=1; for (int x=1;x<=i;x++) { for (int y=i+1;y<=k;y++) { if (calc(way[x-1]+1,way[x],(y==i+1?1:way[y-1]-n+2),way[y]-n+1)!=sum) {u=0;break;} } if (!u) break; } if (u) { isac=1; for (int j=1;j<=k;j++) { if (way[j]<ans[j]) { for (int x=1;x<=k;x++) ans[x]=way[x]; break; } else if (way[j]>ans[j]) break; } } } } if (!isac) printf("Impossible "); else { for (int i=1;i<k;i++) printf("%d ",ans[i]); printf("%d ",ans[k]); } } return 0; //NOTICE LONG LONG!!!!! }
D:打表可知10k~10k+9范围内0~9各出现一次,于是只需要计算零散部分。
#include<bits/stdc++.h> using namespace std; #define ll long long #define inf 1000000010 char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<'0'||c>'9')) c=getchar();return c;} int gcd(int n,int m){return m==0?n:gcd(m,n%m);} ll read() { ll x=0,f=1;char c=getchar(); while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();} while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar(); return x*f; } ll l,r,a[30],f[1000000]; int ff(ll n) { if (n<1000000) return f[n]; int cnt=0; while (n) a[++cnt]=n%10,n/=10; for (int j=1;j<cnt;j++) a[j]+=a[j+1]; a[cnt]=0; ll t=0; for (int j=cnt;j>=1;j--) t=t*10+a[j]%10; return ff(t); } ll calc(ll n) { int s=0;int w=ff(n/10*10); for (ll i=n/10*10;i<=n;i++) s+=(w+(i-n/10*10))%10; return n/10*45+s; } signed main() { for (int i=0;i<=9;i++) f[i]=i; for (int i=10;i<=1000000;i++) { ll cnt=0,x=i; while (x) a[++cnt]=x%10,x/=10; for (int j=1;j<cnt;j++) a[j]+=a[j+1]; a[cnt]=0; ll t=0; for (int j=cnt;j>=1;j--) t=t*10+a[j]%10; f[i]=f[t]; } int T=read(); while (T--) { l=read(),r=read(); cout<<calc(r)-calc(l-1)<<endl; } return 0; //NOTICE LONG LONG!!!!! }
先咕着。