The meaning of problems: statistical interval [a, b] does not contain the numbers 4 and 62 how many.
Ideas: Digital dp
Just can not have a digital 4 can not have consecutive 62, no 4, then at the time of enumeration to determine what, does not enumerate 4 can ensure state legal, so this constraint is not necessary memories of, and for 62, then, involves to two, the current one is not 6 or 6 I count these two different situations are not the same, so use the state to record the number of different schemes.
dp [pos] [sta] represents the current first pos position, whether it is a former 6 states, where sta only two states 0 and 1 to go on it, is not the case 6 can be regarded as the same species, will not affect count.
Code
#include <cstdio> #include <cstring> #include <algorithm> using namespace std; int a[20]; int dp[20][2]; int dfs(int pos, int pre, int sta, int limit){ if (pos==0) return 1; if (!limit&&dp[pos][sta]!=-1) return dp[pos][sta]; int up = limit?a[pos]:9; int tmp = 0; for (int i = 0; i <= up; i++) { if (pre==6&&i==2) continue; if (i==4) continue; tmp += dfs(pos-1, i, i==6, limit&&i==a[pos]); } if(!limit) dp[pos][sta] = tmp; return tmp; } int solve(int x) { int pos = 1; while (x){ a[pos++] = x%10; x /= 10; } return dfs(pos-1, -1, 0, 1); } int main() { int l, r; memset(dp, -1, sizeof(dp)); while (~scanf("%d%d", &l, &r)&&l+r) printf("%d\n", solve(r)-solve(l-1)); return 0; }