[数学 二项式定理 快速幂] BZOJ 3157 && BZOJ 3516 && BZOJ 4126 国王奇遇记

膜拜神犇n+e：http://trinkle.is-programmer.com/2015/6/30/the-adventures-of-the-king.100058.html

本题按时间复杂度的不同共有三种解法。

只会O(m2)的方法

#include<cstdio>#include<cstdlib>#include<algorithm>#define P 1000000007using namespace std;typedef long long ll;const int N=1005;ll n,m,inv,mn,nk;ll C[N][N],f[N];inline ll Pow(ll a,ll b){ll ret=1;for (;b;b>>=1,(a*=a)%=P)if (b&1)(ret*=a)%=P;return ret;}int main(){freopen("t.in","r",stdin);freopen("t.out","w",stdout);scanf("%lld%lld",&n,&m);C[0][0]=1;for (int i=1;i<=m;i++) {C[i][0]=1;for (int j=1;j<=i;j++)(C[i][j]=C[i-1][j]+C[i-1][j-1])%=P;}if(m==1) return printf("%lld\n",(ll)n*(n+1)/2%P),0;mn=Pow(m,n+1);inv=Pow(m-1,P-2);(f[0]=(mn-1)*inv-1)%=P;nk=1;for (int k=1;k<=m;k++) {for(int j=0;j<k;j++)(f[k]+=C[k][j]*((k^j)&1?-1:1)*f[j])%=P;(nk*=n)%=P;f[k]=(nk*mn+f[k])%P*inv%P;}printf("%lld\n",f[m]);return 0;}

本题 m2logn的想法也很不错

#include<cstdio>#include<cstdlib>#include<algorithm>#include<vector>using namespace std;typedef long long ll;typedef vector<ll> Vec;const int P=1e9+7;const int M=200;int n,m;ll C[M+5][M+5];inline void Pre(){  C[0][0]=1;  for (int i=1;i<=M;i++){    C[i][0]=1;    for (int j=1;j<=M;j++)      C[i][j]=(C[i-1][j-1]+C[i-1][j])%P;  }}inline ll Pow(ll a,int b){  ll ret=1;  for (;b;b>>=1,a=a*a%P)    if (b&1)      ret=ret*a%P;  return ret;}inline Vec Solve(int n){  Vec tmp,ret;  ret.resize(m+5);  if (n==1){    for (int i=0;i<=m;i++) ret[i]=m;  }  else if ((n&1) || n==2){    tmp=Solve(n-1);    for (int i=0;i<=m;i++)      ret[i]=tmp[i]+Pow(n,i)*Pow(m,n)%P,ret[i]%=P;  }  else{    tmp=Solve(n/2);    for (int i=0;i<=m;i++){      ll tem=0;      for (int j=0;j<=i;j++)tem+=C[i][j]*Pow(n>>1,i-j)%P*tmp[j],tem%=P;      ret[i]=tmp[i]+tem*Pow(m,n>>1)%P,ret[i]%=P;    }  }  //printf("%d\n",n); for (int i=0;i<=m;i++) printf("%lld ",ret[i]); printf("\n");  return ret;}int main(){  freopen("t.in","r",stdin);  freopen("t.out","w",stdout);  Pre();  scanf("%d%d",&n,&m);  printf("%lld\n",Solve(n)[m]);  return 0;}