#include <cstdio> #include <cstring> #include <cmath> #include <cassert> #include <iostream> #include <algorithm> #include <iterator> #include <string> #include <vector> #include <queue> #include <bitset> #include <utility> #include <stack> #include <numeric> using namespace std; typedef long long LL; typedef vector<int> VI; #define FOR(v,p,k) for(int v=(p);v<=(k);++v) #define FORD(v,p,k) for(int v=(p);v>=(k);--v) #define REP(i,n) for(int i=0;i<(n);++i) #define PB push_back int new_key(int old_key, int digit) { switch( digit ) { case 0: return old_key; case 1: return old_key; case 2: return old_key | 1; case 3: return old_key | 2; case 4: return old_key | 4; case 5: return old_key | 8; case 6: return old_key | 3; case 7: return old_key | 16; case 8: return old_key | 32; case 9: return old_key | 64; default: assert(false); } } const int max_key = 127; const VI digits_for_keys = {2, 3, 4, 5, 7, 8, 9}; bool matches_key(int key, int digit) { return (new_key(0, digit) & key) > 0; } void remove_redundant(VI &digits) { if (count(begin(digits), end(digits), 9)) { digits.erase(remove(begin(digits), end(digits), 3), end(digits)); } if (count(begin(digits), end(digits), 8)) { digits.erase(remove(begin(digits), end(digits), 4), end(digits)); digits.erase(remove(begin(digits), end(digits), 2), end(digits)); } if (count(begin(digits), end(digits), 4)) { digits.erase(remove(begin(digits), end(digits), 2), end(digits)); } } // dp copied from (and changed): https://www.hackerrank.com/topics/digit-dp const int MODULO = 5*7*8*9; LL dp[20][2][max_key+1][MODULO][2];//index (of digit in number), smaller (to control if we can use high digits), key (based on used digits set), modulo remainder, used_zero (already, 0 can be used only at the beginning) void dp_part(const string &s, int index, int smaller, int pow10) { if (index == s.length()) { int key = 0, m = 0, used_zero = 0; dp[index][smaller][key][m][used_zero] = 1; return; } int limit=9; if (smaller) { limit=s[index]-'0'; } FOR(i, 0, limit) { int next_smaller; if(i<s[index]-'0') { next_smaller=0; } else { next_smaller=smaller; } FOR(prev_key, 0, max_key) { int key = new_key(prev_key, i); REP(prev_mod, MODULO) { //for index = s.lenght() - 1 only for prev_mode == 0 we will have non zero entries for index+1 int m = ((pow10 * i) % MODULO + prev_mod) % MODULO; REP(used_zero, 2) { if (i > 0 && used_zero == 0) { dp[index][smaller][key][m][used_zero] += dp[index+1][next_smaller][prev_key][prev_mod][used_zero]; } if (i == 0 && used_zero == 1) { dp[index][smaller][key][m][used_zero] += dp[index+1][next_smaller][prev_key][prev_mod][used_zero]; dp[index][smaller][key][m][used_zero] += dp[index+1][next_smaller][prev_key][prev_mod][1-used_zero]; } } } } } } void dp_solve(LL n) { memset(dp,0,sizeof(dp)); string s=to_string(n); int pow10 = 1; FORD(index, s.length(), 0) { REP(smaller, 2) { dp_part(s, index, smaller, pow10); } //for index == s.lenght() we don't use pow10 if (index < s.length()) { pow10 = (pow10 * 10) % MODULO; } } } LL solve(LL n) { dp_solve(n); LL res = 0; VI digits_used; digits_used.reserve(10); FOR(key, 0, max_key) { digits_used.clear(); for(int d: digits_for_keys) { if (matches_key(key, d)) { digits_used.PB(d); } } remove_redundant(digits_used); int multi = accumulate(begin(digits_used), end(digits_used), 1, multiplies<int>()); REP(i, MODULO/multi) { int m = i*multi; REP(used_zero, 2) { int index = 0, smaller = 1; res += dp[index][smaller][key][m][used_zero]; } } } return res; } int main() { LL l, r; scanf("%lld %lld", &l, &r); printf("%lld\n", solve(r) - solve(l-1)); return 0; }
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 | #include <cstdio> #include <cstring> #include <cmath> #include <cassert> #include <iostream> #include <algorithm> #include <iterator> #include <string> #include <vector> #include <queue> #include <bitset> #include <utility> #include <stack> #include <numeric> using namespace std; typedef long long LL; typedef vector<int> VI; #define FOR(v,p,k) for(int v=(p);v<=(k);++v) #define FORD(v,p,k) for(int v=(p);v>=(k);--v) #define REP(i,n) for(int i=0;i<(n);++i) #define PB push_back int new_key(int old_key, int digit) { switch( digit ) { case 0: return old_key; case 1: return old_key; case 2: return old_key | 1; case 3: return old_key | 2; case 4: return old_key | 4; case 5: return old_key | 8; case 6: return old_key | 3; case 7: return old_key | 16; case 8: return old_key | 32; case 9: return old_key | 64; default: assert(false); } } const int max_key = 127; const VI digits_for_keys = {2, 3, 4, 5, 7, 8, 9}; bool matches_key(int key, int digit) { return (new_key(0, digit) & key) > 0; } void remove_redundant(VI &digits) { if (count(begin(digits), end(digits), 9)) { digits.erase(remove(begin(digits), end(digits), 3), end(digits)); } if (count(begin(digits), end(digits), 8)) { digits.erase(remove(begin(digits), end(digits), 4), end(digits)); digits.erase(remove(begin(digits), end(digits), 2), end(digits)); } if (count(begin(digits), end(digits), 4)) { digits.erase(remove(begin(digits), end(digits), 2), end(digits)); } } // dp copied from (and changed): https://www.hackerrank.com/topics/digit-dp const int MODULO = 5*7*8*9; LL dp[20][2][max_key+1][MODULO][2];//index (of digit in number), smaller (to control if we can use high digits), key (based on used digits set), modulo remainder, used_zero (already, 0 can be used only at the beginning) void dp_part(const string &s, int index, int smaller, int pow10) { if (index == s.length()) { int key = 0, m = 0, used_zero = 0; dp[index][smaller][key][m][used_zero] = 1; return; } int limit=9; if (smaller) { limit=s[index]-'0'; } FOR(i, 0, limit) { int next_smaller; if(i<s[index]-'0') { next_smaller=0; } else { next_smaller=smaller; } FOR(prev_key, 0, max_key) { int key = new_key(prev_key, i); REP(prev_mod, MODULO) { //for index = s.lenght() - 1 only for prev_mode == 0 we will have non zero entries for index+1 int m = ((pow10 * i) % MODULO + prev_mod) % MODULO; REP(used_zero, 2) { if (i > 0 && used_zero == 0) { dp[index][smaller][key][m][used_zero] += dp[index+1][next_smaller][prev_key][prev_mod][used_zero]; } if (i == 0 && used_zero == 1) { dp[index][smaller][key][m][used_zero] += dp[index+1][next_smaller][prev_key][prev_mod][used_zero]; dp[index][smaller][key][m][used_zero] += dp[index+1][next_smaller][prev_key][prev_mod][1-used_zero]; } } } } } } void dp_solve(LL n) { memset(dp,0,sizeof(dp)); string s=to_string(n); int pow10 = 1; FORD(index, s.length(), 0) { REP(smaller, 2) { dp_part(s, index, smaller, pow10); } //for index == s.lenght() we don't use pow10 if (index < s.length()) { pow10 = (pow10 * 10) % MODULO; } } } LL solve(LL n) { dp_solve(n); LL res = 0; VI digits_used; digits_used.reserve(10); FOR(key, 0, max_key) { digits_used.clear(); for(int d: digits_for_keys) { if (matches_key(key, d)) { digits_used.PB(d); } } remove_redundant(digits_used); int multi = accumulate(begin(digits_used), end(digits_used), 1, multiplies<int>()); REP(i, MODULO/multi) { int m = i*multi; REP(used_zero, 2) { int index = 0, smaller = 1; res += dp[index][smaller][key][m][used_zero]; } } } return res; } int main() { LL l, r; scanf("%lld %lld", &l, &r); printf("%lld\n", solve(r) - solve(l-1)); return 0; } |