M-SOLUTIONS Programming Contest

A.(n-2)*180

 #include<cstdio>
 #include<cstring>
 #include<iostream>
 #include<algorithm>
 #define rep(i,l,r) for (int i=(l); i<=(r); i++)
 typedef long long ll;
 using namespace std;

 int n;

 int main(){
     scanf("%d",&n); printf("%d\n",(n-)*);
     return ;
 }

B.已确定的输局<=7

 #include<cstdio>
 #include<cstring>
 #include<iostream>
 #include<algorithm>
 #define rep(i,l,r) for (int i=(l); i<=(r); i++)
 typedef long long ll;
 using namespace std;

 int n,x;
 char s[];

 int main(){
     scanf("%s",s+); n=strlen(s+);
     rep(i,,n) if (s[i]=='x') x++;
     if (x>) puts("NO"); else puts("YES");
     return ;
 }

C.现只考虑A最后胜的情况，B同理。枚举A赢第n局之前B赢了多少局i，那么若不考虑平局概率则这种情况的发生概率为A^n*B^i*C(n-1+i,i)，由期望显然可以得到，一场非平局的出现概率为1-C则期望1/(1-C)会出现一场非平局，共有n+i个非平局，则期望局数为(n+i)/(1-C)。

 #include<cstdio>
 #include<cstring>
 #include<iostream>
 #include<algorithm>
 #define rep(i,l,r) for (int i=(l); i<=(r); i++)
 typedef long long ll;
 using namespace std;

 const int N=,mod=1e9+;
 int n,A,B,C,ans,fac[N],inv[N];

 int ksm(int a,int b){
     int res=;
     for (; b; a=1ll*a*a%mod,b>>=)
         if (b & ) res=1ll*res*a%mod;
     return res;
 }

 int cc(int n,int m){ return n<m ?  : 1ll*fac[n]*inv[m]%mod*inv[n-m]%mod; }

 int main(){
     scanf("%d%d%d%d",&n,&A,&B,&C); int ii=ksm(,mod-);
     A=1ll*A*ii%mod; B=1ll*B*ii%mod; C=1ll*C*ii%mod;
     int a=1ll*ksm(A+B,mod-)*A%mod,b=1ll*ksm(A+B,mod-)*B%mod;
     fac[]=; rep(i,,n+n) fac[i]=1ll*fac[i-]*i%mod;
     inv[n+n]=ksm(fac[n+n],mod-); for (int i=n+n; i; i--) inv[i-]=1ll*inv[i]*i%mod;
     rep(i,,n-){
         ans=(ans+1ll*ksm(b,i)*ksm(a,n)%mod*cc(n-+i,i)%mod*ksm(-C+mod,mod-)%mod*(n+i))%mod;
         ans=(ans+1ll*ksm(a,i)*ksm(b,n)%mod*cc(n-+i,i)%mod*ksm(-C+mod,mod-)%mod*(n+i))%mod;
     }
     printf("%d\n",ans);
     return ;
 }

D.能取到最大值=总和-最大点权。于是把c从大到小排序，按DFS序分配即可。

 #include<cstdio>
 #include<cstring>
 #include<iostream>
 #include<algorithm>
 #define rep(i,l,r) for (int i=(l); i<=(r); i++)
 #define For(i,x) for (int i=h[x],k; i; i=nxt[i])
 typedef long long ll;
 using namespace std;

 const int N=;
 int n,u,v,tot,ans,cnt,c[N],s[N],h[N],to[N<<],nxt[N<<];
 struct E{ int u,v; }e[N];
 void add(int u,int v){ to[++cnt]=v; nxt[cnt]=h[u]; h[u]=cnt; }
 bool cmp(int a,int b){ return a>b; }

 void dfs(int x,int fa){
     s[x]=c[++tot];
     For(i,x) if ((k=to[i])!=fa) dfs(k,x);
 }

 int main(){
     scanf("%d",&n);
     rep(i,,n) scanf("%d%d",&u,&v),e[i]=(E){u,v},add(u,v),add(v,u);
     rep(i,,n) scanf("%d",&c[i]);
     sort(c+,c+n+,cmp); dfs(,);
     rep(i,,n) ans+=min(s[e[i].u],s[e[i].v]);
     printf("%d\n",ans);
     rep(i,,n) printf("%d ",s[i]);
     return ;
 }

E.答案等于(x/d)*(x/d+1)*...*(x/d+(n-1))*d^n，这就是个阶乘乘上快速幂。注意特判d=0或x/d~x/d+n-1中出现0的情况。

 #include<cstdio>
 #include<algorithm>
 #define rep(i,l,r) for (int i=(l); i<=(r); i++)
 using namespace std;

 const int mod=1e6+;
 int x,d,n,T,fac[mod+];

 int ksm(int a,int b){
     int res=;
     for (; b; a=1ll*a*a%mod,b>>=)
         if (b & ) res=1ll*res*a%mod;
     return res;
 }

 int main(){
     fac[]=; rep(i,,mod-) fac[i]=1ll*fac[i-]*i%mod;
     for (scanf("%d",&T); T--; ){
         scanf("%d%d%d",&x,&d,&n);
         if (!d){ printf("%d\n",ksm(x,n)); continue; }
         x=1ll*x*ksm(d,mod-)%mod;
         if (!x || x+n->=mod) puts("");
             else printf("%lld\n",1ll*fac[x+n-]*ksm(fac[x-],mod-)%mod*ksm(d,n)%mod);
     }
     return ;
 }