sign。。。很难的数位DP。。。。。
#include <iostream> #include <queue> #include <stack> #include <map> #include <set> #include <bitset> #include <cstdio> #include <algorithm> #include <cstring> #include <climits> #include <cstdlib> #include <cmath> #include <time.h> #define maxn 1005 #define maxm 200005 #define eps 1e-10 #define mod 1000000007 #define INF 0x3f3f3f3f #define PI (acos(-1.0)) #define lowbit(x) (x&(-x)) #define mp make_pair #define ls o<<1 #define rs o<<1 | 1 #define lson o<<1, L, mid #define rson o<<1 | 1, mid+1, R //#pragma comment(linker, "/STACK:16777216") typedef long long LL; typedef unsigned long long ULL; //typedef int LL; using namespace std; LL qpow(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base;base=base*base;b/=2;}return res;} LL powmod(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base%mod;base=base*base%mod;b/=2;}return res;} void scanf(int &__x){__x=0;char __ch=getchar();while(__ch==' '||__ch=='\n')__ch=getchar();while(__ch>='0'&&__ch<='9')__x=__x*10+__ch-'0',__ch = getchar();} LL gcd(LL _a, LL _b){if(!_b) return _a;else return gcd(_b, _a%_b);} // head const int tt = 2520; LL dp[20][3000][50]; int x[50], mpp[3000]; int s[50]; int cnt; void init(void) { cnt = 0; for(int i = 1; i <= tt; i++) if(tt % i == 0) x[++cnt] = i, mpp[i] = cnt; memset(dp, -1, sizeof dp); } LL dfs(int pos, int sum, int lcm, int limit) { if(pos == 0) { if(sum % x[lcm] == 0) return 1; else return 0; } if(dp[pos][sum][lcm] != -1 && !limit) return dp[pos][sum][lcm]; lcm = x[lcm]; int b = (limit ? s[pos] : 9); LL ans = 0; for(int i = 0; i <= b; i++) { int nlcm; if(!i) nlcm = lcm; else nlcm = lcm / gcd(lcm, i) * i; ans += dfs(pos - 1, (sum * 10 + i) % tt, mpp[nlcm], limit && i == b); } if(!limit) dp[pos][sum][mpp[lcm]] = ans; return ans; } LL calc(LL t) { int size = 0; while(t) { s[++size] = t % 10; t /= 10; } return dfs(size, 0, 1, 1); } int main(void) { int _; LL a, b; init(); while(scanf("%d", &_)!=EOF) { while(_--) { scanf("%I64d%I64d", &a, &b); printf("%I64d\n", calc(b) - calc(a - 1LL)); } } return 0; }