注意处理数字只有一位的情况(其实不用怎么处理)= =
简单数位DP
#include <cstdio> #include <cstring> #include <cmath> #include <algorithm> #include <climits> #include <string> #include <iostream> #include <map> #include <cstdlib> #include <list> #include <set> #include <queue> #include <stack> using namespace std; typedef long long LL; const int maxn = 20; int lim[maxn],len; int num[maxn]; int f[15][15][15][2]; void getlim(int num) { memset(lim,0,sizeof(lim)); len = 0; while(num) { lim[len++] = num % 10; num /= 10; } } int myabs(int x) { return x < 0 ? -x : x; } int dfs(int now,int prev,int maxd,int first,int bound) { int ¬e = f[now][prev][maxd][first]; if(now == 0) { if(first && maxd >= 2) { return 1; } return 0; } if(!bound && note != -1) return note; int m = bound ? lim[now - 1] : 9,ret = 0; for(int i = 0;i <= m;i++) { num[now - 1] = i; if(first) { int nmaxd = min(maxd,myabs(i - prev)); ret += dfs(now - 1,i,nmaxd,1,bound && i == m); } else { ret += dfs(now - 1,i,maxd,i || first,bound && i == m); } } if(!bound) note = ret; return ret; } int work(int num) { memset(f,-1,sizeof(f)); if(num < 10) return num; getlim(num); return dfs(len,0,11,0,1); } int main() { int n,m; while(scanf("%d%d",&n,&m) == 2) { printf("%d\n",work(m) - work(n - 1)); } return 0; }