[DP] 51nod1048 integer decomposed to a power V2__DP of 2

This is an upgraded version of 51nod1383. n N is 1030 10^{30}
The approach is a fantastic DP DP. Consider under the binary system.
Fi,j F_{i,j} indicates that the maximum number of 2i 2^i is the number of scenarios when 2j 2^j.
Transfer is fi,j=fi−1,k+fi−1−k,j−k f_{i,j}=f_{i-1,k}+f_{i-1-k,j-k}
Then there is a similar merge per bit. Gi,j G_{i,j} represents the number of scenarios in which the maximum number of 2j 2^j is composed of n n binary bits. Transfer similar.
God TM does not give modulus to force high precision ... and ...

The following is an Kathang code

#include <cstdio> #include <cstring> using namespace std;
typedef long Long LL;
const int maxn=101,con=100000000;
    inline int max (int x,int y) {return x>y?x:y;} struct int{int a[205];
    inline int (int val=0) {memset (a,0,sizeof (a)); doing A[++a[0]]=val%con, Val/=con; while (Val);}
        inline void Sread () {static char st[6000]; scanf ("%s", st+1); int Len=strlen (st+1); a[0]= (len+7)/8;
    for (int i=1;i<=len;i++) a[(len+1-i+7)/8]=a[(len+1-i+7)/8]*10+st[i]-' 0 ';
    } inline void Print () {printf ("%d", a[a[0]]); for (int i=a[0]-1;i>=1;i--) printf ("%08d", A[i);}
        inline void operator + = (const Int &b) {A[0]=max (a[0],b.a[0]);
        for (int i=1;i<=a[0];i++) a[i]+=b.a[i], A[i+1]+=a[i]/con, A[i]%=con;
    if (a[a[0]+1]) a[0]++; } Inline int operator * (const int &b) {Int C; c.a[0]=a[0]+b.a[0]-1;
        LL v; for (int i=1;i<=a[0];i++) for (int j=1;j<=b.a[0];j++) v= (LL) a[i]*b.a[j], c.a[i+j]+= (v+c.a[i+j-1))/Con, c.a[i+j-1]= (v+c.a[i+j-1])%con; if (c.a[c.a[0]+1]) c.a[0]++; while (C.a[0]>1&&!c.a[c.a[0]]) c.a[0]--;
    return C; } inline void operator/= (const int val) {LL x=0; for (int i=a[0];i>=1;i--) x=x*con+a[i], A[i]=x/val, x=x%v
        Al
    while (A[0]>1&&!a[a[0]]) a[0]--;
}} N,f[maxn][maxn],g[2][maxn],ans;
    int main () {n.sread ();
    F[0][0]=1;
        for (int i=1;i<=100;i++) {f[i][i]=1;
    for (int j=0;j<=i-1;j++) for (int k=0;k<=j;k++) f[i][j]+=f[i-1][k]*f[i-1-k][j-k];
    int now=0;
        for (int i=0;i<=100;n/=2,i++) if (n.a[1]&1) {now++;
            if (now==1) {for (int j=0;j<=i;j++) g[now&1][j]=f[i][j];
        Continue
            for (int j=0;j<=i;j++) {g[now&1][j]=0;
        for (int k=0;k<=j;k++) g[now&1][j]+=g[(now&1) ^1][k]*f[i-k][j-k];
    for (int i=0;i<=100;i++) ans+=g[now&1][i]; Ans.
    Print (); REturn 0; }