数位统计DP-计数问题
分情况讨论:【a, b】,0~9
count(n, x),1~n中x出现的次数。
【a, b】中x出现的次数, 用前缀和:count(b, x) - count(a-1, x);
1 ~ n,x = 1
n = abcdefg 分别求出1在每一位上出现的次数
eg:比如求1在第4位(d)上出现的次数?
1 <= xxx1yyy<=abcdefg
(1) xxx = 000 ~ abc-1, yyy = 000~999, abcd * 1000
(2) xxx = abc
(2.1) d < 1, abc1yyy > abc0efg, 0
(2.2) d = 1, efg + 1;
(2.3 d > 1, yyy = 000 ~ 999, 1000
如果x = 0,那么001 ~ abc - 1
#include <iostream> #include <algorithm> #include <vector> using namespace std; const int N = 10; /* 001~abc-1, 999 abc 1. num[i] < x, 0 2. num[i] == x, 0~efg 3. num[i] > x, 0~999 */ int get(vector<int> num, int l, int r) { int res = 0; for (int i = l; i >= r; i -- ) res = res * 10 + num[i]; return res; } int power10(int x) { int res = 1; while (x -- ) res *= 10; return res; } int count(int n, int x) { if (!n) return 0; vector<int> num; while (n) { num.push_back(n % 10); n /= 10; } n = num.size(); int res = 0; for (int i = n - 1 - !x; i >= 0; i -- ) { if (i < n - 1) { res += get(num, n - 1, i + 1) * power10(i); if (!x) res -= power10(i); } if (num[i] == x) res += get(num, i - 1, 0) + 1; else if (num[i] > x) res += power10(i); } return res; } int main() { int a, b; while (cin >> a >> b , a) { if (a > b) swap(a, b); for (int i = 0; i <= 9; i ++ ) cout << count(b, i) - count(a - 1, i) << ' '; cout << endl; } return 0; }