简单的数位DP 刚开始全以2进制来算的 后来发现要找最接近x,y值的那个基于b进制的0,1组合
1 #include <iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<algorithm> 5 #include<stdlib.h> 6 #include<cmath> 7 using namespace std; 8 #define LL __int64 9 #define INF 1e11 10 LL dp[35][25],pp[12][22]; 11 int g1,g2,p[1<<21]; 12 void init() 13 { 14 int i,j; 15 for(i = 3 ; i <= 10 ; i++) 16 { 17 pp[i][0] = 1; 18 for(j = 1 ; j <= 20 ; j++) 19 { 20 pp[i][j] = pp[i][j-1]*i; 21 if(pp[i][j]>INF) 22 break; 23 } 24 } 25 dp[0][0] = 1; 26 dp[1][1] = 1; 27 dp[1][0] = 1; 28 for(i = 2; i <= 32 ; i++) 29 { 30 dp[i][0] = 1; 31 for(j =1; j <= 20 ; j++) 32 dp[i][j] = dp[i-1][j-1]+dp[i-1][j]; 33 } 34 } 35 void change(int x,int y,int b,int c1[],int c2[]) 36 { 37 int di[35],g=0,i; 38 while(x) 39 { 40 di[++g] = x%b; 41 x/=b; 42 } 43 for(i = g ; i >= 1;i--) 44 c1[i] = di[i]; 45 g1 = g; 46 g = 0; 47 while(y) 48 { 49 di[++g] = y%b; 50 y/=b; 51 } 52 for(i = g ; i >= 1;i--) 53 c2[i] = di[i]; 54 g2 = g; 55 } 56 LL find(int c[],int k,int len) 57 { 58 int i,j,o=0; 59 LL sum=0; 60 for(i = len ; i >= 1 ; i--) 61 { 62 if(c[i]==0) continue; 63 if(k>=o) 64 sum+=dp[i-1][k-o]; 65 o++; 66 } 67 if(o==k) 68 sum++; 69 return sum; 70 } 71 int main() 72 { 73 int x,y,k,b,j,i; 74 int c1[35],c2[35]; 75 init(); 76 scanf("%d%d%d%d",&x,&y,&k,&b); 77 if(b!=2) 78 { 79 LL s=0,x1=-1,x2=-1; 80 x--; 81 for(i = 0 ; i <= (1<<20) ; i++) 82 { 83 s =0 ; 84 for(j = 0 ; j <= 20 ; j++) 85 if(i&(1<<j)) 86 { 87 s+=pp[b][j]; 88 } 89 if(x1==-1&&s>x) 90 x1 = i-1; 91 if(s>y) 92 { 93 x2 = i-1; 94 break; 95 } 96 } 97 for(j = 20 ; j >=0 ; j--) 98 if(x1&(1<<j)) 99 { 100 c1[j+1] = 1; 101 if(!g1) 102 g1 = j+1; 103 } 104 else 105 c1[j+1] = 0; 106 for(j = 20 ; j >=0 ; j--) 107 if(x2&(1<<j)) 108 { 109 c2[j+1] = 1; 110 if(!g2) 111 g2 = j+1; 112 } 113 else 114 c2[j+1] = 0; 115 } 116 else 117 change(x-1,y,b,c1,c2); 118 LL s1 = find(c1,k,g1); 119 LL s2 = find(c2,k,g2); 120 printf("%I64d ",s2-s1); 121 return 0; 122 }