[Bzoj 3209] The number of flowers and gods

A simple digital DP question

But there is only _ builtin_popcountll in my mind (self-respect)

I quickly understood the problem after reading the answer, and there was such a simple question that I did not expect the practice ~ Feeling

 

F [I] indicates the number of digits with 1 being I and less than N.

Then the answer is explain if [I], and f [I] can be messed up from high to low DP with a combination of numbers. O (lgn) 2)

For example, the front I-1 bit has K-bit 1, the first I-bit is 1, followed by J-bit

Then, set the I-th bit to 0. No matter what number the last J-th bit is, it is smaller than N, use the number of combinations to update f [K + L] (0 <= L <= J ).

 

However, after wa for a long time, we found that the modules commonly used in the 10000007 competition were

No! Yes! Quality! Count!

10000007 = 941*10627

Then I used the ferma's Theorem to do this, biubiu ~

You have to calculate the answer by enumeration.

Pay attention to it later.

 

 1 #include <cstdio> 2 typedef long long LL; 3 const int mod=10000007; 4 const int size=64; 5  6 LL n; 7 LL ans; 8 LL c[size][size]; 9 LL f[1024];10 inline LL getint();11 inline void putint(LL);12 inline int lg(LL);13 inline LL mul(LL, LL);14 inline LL pow(LL, LL);15 inline void calc();16 17 int main()18 {19     n=getint();20     ans=1;21     calc();22     putint(ans);23 24     return 0;25 }26 inline LL getint()27 {28     register LL num=0;29     register char ch;30     do ch=getchar(); while (ch<‘0‘ || ch>‘9‘);31     do num=num*10+ch-‘0‘, ch=getchar(); while (ch>=‘0‘ && ch<=‘9‘);32     return num;33 }34 inline void putint(LL num)35 {36     char stack[21];37     register int top=0;38     if (num==0) stack[top=1]=‘0‘;39     for ( ;num;num/=10) stack[++top]=num%10+‘0‘;40     for ( ;top;top--) putchar(stack[top]);41     putchar(‘\n‘);42 }43 inline int lg(LL x)44 {45     int ret=0;46     for ( ;x>1;x>>=1) ret++;47     return ret;48 }49 inline LL mul(LL a, LL b)50 {51     return a*b%mod;52 }53 inline LL pow(LL a, LL b)54 {55     LL c=1;56     for ( ;b;b>>=1)57     {58         if (b & 1) c=mul(c, a);59         a=mul(a, a);60     }61     return c;62 }63 inline void calc()64 {65     for (int i=0;i<64;i++)66     {67         c[i][0]=1;68         for (int j=1;j<i;j++)69             c[i][j]=c[i-1][j-1]+c[i-1][j];70         c[i][i]=1;71     }72     int k=0;73     for (int i=lg(n);i>=0;i--) if ((n>>i) & 1)74     {75         ans=mul(ans, k+1);76         for (int j=1;j<=i;j++)77             ans=mul(ans, pow(k+j, c[i][j]));78         ++k;79     }80 }

Ben silly wa Outang Series

 

[Bzoj 3209] The number of flowers and gods