# Codeforces 300-A/B/C

Source: Internet
Author: User

Codeforces 300-A/B/C

Question A is A water question. Just classify it as required.

``````#include

#include

#include

using namespace std;int main(){    int n,cnt1,cnt2,cnt3,a[110];    int b1[110],b2[110],b3[110];    while(scanf("%d",&n)!=EOF)    {        cnt1=cnt2=cnt3=0;        for(int i=0;i

0)            {                b2[cnt2++]=x;            }         }            printf("1 %d\n",b1[0]);            if(cnt2==0)            {                b2[cnt2++]=b3[--cnt3];                b2[cnt2++]=b3[--cnt3];            }            printf("%d ",cnt2);            for(int i=0;i<cnt2;i++) printf("%d="" ",b2[i]);="" printf("\n%d="" 0="" ",cnt3+1);="" for(int="" i="0;i

``````

`In question B, we should first mark the number of people in a team as the same number. Then, we will randomly divide the team into several groups (TAGS) for those who do not have Team requirements ); then, let's check whether the number of teams is greater than three or less than three is impossible. Output-1. Other situations are marked as output teams. Code writing is messy.`

`The Code is as follows:`

``#include  #include  #include  using namespace std;int M[50][50],vis[50];int m,n;int dfs(int x,int cnt){ vis[x]=cnt; for(int i=1;i<=n;i++) if(M[x][i]) { M[x][i]=M[i][x]=0; dfs(i,cnt); }}int main(){ while(scanf("%d%d",&n,&m)!=EOF) { memset(vis,0,sizeof(vis)); memset(M,0,sizeof(M)); for(int i=1;i<=m;i++) { int a,b; scanf("%d%d",&a,&b); if(a==b) continue; M[a][b]=1; M[b][a]=1; } int flag=0,cnt=0; for(int i=1;i<=n;i++) { int j; for(j=1;j<=n;j++) { if(M[i][j]) break; } if(j<=n) dfs(i,++cnt); } for(int i=1;i<=n/3;i++) { int d=0; for(int j=1;j<=n;j++) if(vis[j]==i) d++; if(d>3) { flag=1; break; } while(d<3) { for(int k=1;k<=n;k++) if(vis[k]==0) { vis[k]=i; break; } d++; } } for(int i=1;i<=n/3;i++) { int d=0; for(int j=1;j<=n;j++) { if(vis[j]==i) d++; } if(d<3) { flag=12; break; } } if(flag) { printf("-1\n"); continue; } for(int i=1;i<=n/3;i++) { for(int j=1;j<=n;j++) { if(vis[j]==i) printf("%d ",j); } printf("\n"); } } return 0;}   ``

`In question C, we should enumerate (construct) the result and check whether it meets the requirements. Then there is a problem of arrangement and combination.Note a formula: (a/B) % mod = a * pow (B, mod-2). mod is a prime number. This formula can be used to quickly calculate the number of combinations.`

`The Code is as follows:`

``#include  #include  #include  using namespace std;#define mod 1000000007#define ll long longll f[1000009];ll pow(ll a,ll b){ ll ans=1; while(b) { if(b&1) ans=ans*a%mod; b>>=1; a=a*a%mod; } return ans%mod;}int a,b,n;bool check(ll x){ while(x) { int i=x%10; if(i!=a&&i!=b) return false; x/=10; } return true;}int main(){ f[0]=1; for(int i=1;i<=1000001;i++) f[i]=f[i-1]*i%mod; while(scanf("%d%d%d",&a,&b,&n)!=EOF) { ll ans=0; for(int i=0;i<=n;i++) { ll sum=(ll)i*a+(ll)(n-i)*b; if(check(sum)) { ll d=f[i]*f[n-i]%mod; ans=(ans+f[n]*pow(d,mod-2)%mod)%mod; } } printf("%lld\n",ans); } return 0;}   ``

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