C 纸牌游戏
涨薪
https://ac.nowcoder.com/acm/contest/7606/C
C 纸牌游戏
时,这时的
的还没被淘汰,所以还得加上
时,这时
的已经被淘汰了,所以降序排序后,就一直取前
的数即可,快速幂写一下就行了。
#include<bits/stdc++.h> #pragma GCC optimize(2) #pragma GCC optimize(3) #pragma GCC optimize("Ofast") #pragma GCC target("avx,avx2,fma") #pragma GCC optimization("unroll-loops") using namespace std; #define IOS ios::sync_with_stdio(false);cin.tie(0);cout.tie(0); #define ll long long #define ull unsigned long long #define rint register int #define ld long double #define db double #define rep(i, l, r) for (rint i = l; i <= r; i++) #define rep1(i, a, n) for (rint i = a; i < n; i++) #define per(i, l, r) for (rint i = l; i >= r; i--) #define per1(i ,a, n) for (rint i = a; i > n; i--) #define mset(s, _) memset(s, _, sizeof(s)) #define pb push_back #define pii pair <int, int> #define mp(a, b) make_pair(a, b) #define sd(x) scanf("%d",&(x)) #define slld(x) scanf("%lld",&(x)) #define sdd(x,y) scanf("%d%d",&(x),&(y)) #define sc(s) scanf("%s",(s)) #define pd(x) printf("%d\n",(x)) #define plld(x) printf("%lld\n",(x)) #define pdk(x) printf("%d ",(x)) const int inf=0x3f3f3f3f; namespace IO{ char ibuf[1<<21],*ip=ibuf,*ip_=ibuf; char obuf[1<<21],*op=obuf,*op_=obuf+(1<<21); inline char gc(){ if(ip!=ip_)return *ip++; ip=ibuf;ip_=ip+fread(ibuf,1,1<<21,stdin); return ip==ip_?EOF:*ip++; } inline void pc(char c){ if(op==op_)fwrite(obuf,1,1<<21,stdout),op=obuf; *op++=c; } inline ll read(){ register ll x=0,ch=gc(),w=1; for(;ch<'0'||ch>'9';ch=gc())if(ch=='-')w=-1; for(;ch>='0'&&ch<='9';ch=gc())x=x*10+ch-48; return w*x; } template<class I> inline void write(I x){ if(x<0)pc('-'),x=-x; if(x>9)write(x/10);pc(x%10+'0'); } class flusher_{ public: ~flusher_(){if(op!=obuf)fwrite(obuf,1,op-obuf,stdout);} }IO_flusher; } using namespace IO; ll qm (ll a, ll b ,ll c){ ll ret=1%c; while(b){ if(b&1) ret=ret*a%c; a=a*a%c; b=b>>1; } return ret; } const int N=1e6+10; const int mod=1e9+7; ll n,m,x,y,a[N],ans1,ans2,ans; int main() { n=read();m=read();x=read();y=read(); rep(i, 1, n) { a[i]=read(); } sort(a+1,a+1+n,greater<ll>()); rep(i, 1, n){ if(i<=x) ans1=(ans1+a[i])%mod; if(i>x && i<=x+y) ans2=(ans2+a[i])%mod; } if(m < 2){ rep(i, x+y+1, n){ ans=(ans+a[i])%mod; } } write((ans%mod+ans1*qm(3,m,mod)%mod+ans2*qm(2,m,mod)%mod)%mod); return 0; }