codeforces 286E Ladies' Shop

题目大意：n个小于等于m的数，现在你需要在[1,m]中选择若干个数，使得选出的数能组成的所有数正好与n个数相同，给出最少要选多少个数。

题目分析：

结论一：选择的若干个数一定在n个数中。

证明：否则的话不满足“正好”。

结论二：若a,b在由n个数组成的集合中，则a+b(a+b<=m)也在由n个数组成的集合中。

证明：通过归纳法可以证明。

那么我们考虑构造生成函数G(x)=∑ki*xⁱ,其中当由n组成的集合中有数i时ki=1，否则为0。接着将多出的数删除即可。

代码：

#include<bits/stdc++.h>
using namespace std;typedef long long ll;const ll mod = 998244353;
const int gg = 3;
const int maxn = 2520210;
int mp[maxn],a[maxn],f[maxn];
int n,m;
int ord[maxn];int fast_pow(int now,int p){if(p == 0) return 1;if(p == 1) return now;int z = fast_pow(now,p/2); z = (1ll*z*z)%mod;if(p & 1){z = (1ll*z*now)%mod;}return z;
}void fft(int *d,int len,int kind){for(int i=0;i<len;i++) if(ord[i] > i) swap(d[i],d[ord[i]]);for(int i=1;i<len;i<<=1){int wn = fast_pow(gg,(mod-1)/(i<<1));if(kind == -1) wn = fast_pow(wn,mod-2);for(int j=0;j<len;j += (i<<1)){int w = 1;for(int k=0;k<i;k++,w=(1ll*w*wn)%mod){ll x = d[j+k],y = (1ll*w*d[j+k+i])%mod;d[j+k] = (x+y)%mod; d[j+k+i] = (x-y+mod)%mod;}}}if(kind == -1){int inv = fast_pow(len,mod-2);for(int i=0;i<len;i++) d[i] = (1ll*d[i]*inv)%mod;}
}void read(){scanf("%d%d",&n,&m);for(int i=1;i<=n;i++) scanf("%d",&a[i]),mp[a[i]] = f[a[i]] = 1;
}void work(){int bit = 0,len = 1;while(len < m*2) bit++,len<<=1;for(int i=1;i<len;i++) ord[i] = (ord[i>>1]>>1) + ((i&1)<<bit-1);fft(f,len,1);for(int i=0;i<len;i++) f[i] = (1ll*f[i]*f[i])%mod;fft(f,len,-1);int ans = 0;for(int i=0;i<=m;i++){if(f[i]&&mp[i]) {mp[i] = 0;continue;}if(f[i]){puts("NO");return;}if(mp[i]) ans++;}puts("YES");printf("%d\n",ans);for(int i=0;i<=m;i++) if(mp[i]) printf("%d ",i);
}int main(){read();work();return 0;
}

转载于:https://www.cnblogs.com/1-1-1-1/p/8572193.html

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