论文：
http://hi.baidu.com/3xianbin/item/917aca907a3fb6f4291647fc
http://wenku.baidu.com/view/d2414ffe04a1b0717fd5dda8.html
其中一道题目求给定区间的所有数的k进制的各位数字的和：
#include <iostream> #include <cstdio> #include <cmath> #include <vector> #include <cstring> #include <algorithm> #include <string> #include <set> #include <functional> #include <numeric> #include <sstream> #include <stack> #include <map> #include <queue> #define CL(arr, val) memset(arr, val, sizeof(arr)) #define lc l,m,rt<<1 #define rc m + 1,r,rt<<1|1 #define pi acos(-1.0) #define ll __int64 #define L(x) (x) << 1 #define R(x) (x) << 1 | 1 #define MID(l, r) (l + r) >> 1 #define Min(x, y) (x) < (y) ? (x) : (y) #define Max(x, y) (x) < (y) ? (y) : (x) #define E(x) (1 << (x)) #define iabs(x) (x) < 0 ? -(x) : (x) #define OUT(x) printf("%I64d\n", x) #define lowbit(x) (x)&(-x) #define Read() freopen("din.txt", "r", stdin) #define Write() freopen("dout.txt", "w", stdout); #define M 137 #define N 207 using namespace std; const int inf = 0x7f7f7f7f; const int mod = 1000000007; ll x,y,n,k; ll getsum(ll pre,ll n,ll k) { ll C = pre, B = 1; for (int i = 0; i < n; ++i) { B *= k; C *= k; } B = B*n*(k - 1)/2; return B + C; } ll dfs(ll pre,ll n,ll k) { ll res = 0; if (n < k) { for (int i = 0; i <= n; ++i) { res += pre + i; } return res; } int d = 0; ll t = 1; ll tn = n; while (tn >= k) { tn /= k; t *= k; d++; } for (int i = 0; i < tn; ++i) { res += getsum(pre + i,d,k); } res += dfs(pre + tn,n - tn*t,k); return res; } int main() { while (cin>>x>>y>>k) { ll sum1 = dfs(0,y,k); ll sum2 = dfs(0,x - 1,k); cout<<sum1<<" "<<sum2<<" "<<sum1 - sum2<<endl; } return 0; }