#include <cstdio> #include <iostream> #include <algorithm> #include <cstring> #include <vector> #include <set> #include <map> #include <cmath> #include <list> #include <ctime> #include <sstream> #include <queue> #include <stack> #include <bitset> #include <unordered_set> #include <unordered_map> using namespace std; typedef vector<int> vi; typedef pair<int,int> pii; typedef long long ll; typedef short int sint; #define FOR(x, b, e) for(int x=(b); x<=(e); ++x) #define FORD(x, b, e) for(int x=((int)(b))-1; x>=(e); --x) #define REP(x, n) for(int x=0; x<(n); ++x) #define ALL(c) c.begin(),c.end() #define SIZE(x) ((int)((x).size())) #define PB push_back #define ST first #define ND second #define mp(x,y) make_pair(x,y) #define DEBUG 1 #define debug(x) {if (DEBUG)cerr <<#x <<" = " <<x <<endl; } #define debugv(x) {if (DEBUG) {cerr <<#x <<" = "; FOREACH(it, (x)) cerr <<*it <<", "; cout <<endl; }} #define REMAX(a,b) (a)=max((a),(b)); #define REMIN(a,b) (a)=min((a),(b)); #define wez(n) int (n); scanf("%d",&(n)); #define wez2(n,m) int (n),(m); scanf("%d %d",&(n),&(m)); const int N = 202010; int tab[N]; int n; string toString(int x) { ostringstream s; s << x; return s.str(); } struct Num { string pref; // first 10 chars int nastLen; // length after pref, if pref has 10 chars int suffix; // the ending numbers of the suffix, helpful to check if we can do suffix+1 != 10^suffLen Num(string pref_, int suffLen_ = 0, int suffix_ = 0): pref(pref_), nastLen(suffLen_), suffix(suffix_) { } int size() const { return nastLen + pref.size(); } }; bool canIncreaseSuffix(Num a) { return (a.nastLen > 6 || pow(10, a.nastLen) > a.suffix + 1); } bool canIncreaseString(string x) { REP(i, x.size()) { if (x[i] != '9') { return true; } } return false; } Num addZeros(string nextStringNum, int dod) { while (dod > 0 && nextStringNum.size() < 10) { nextStringNum += '0'; --dod; } return Num(nextStringNum, dod, 0); } pair<int, Num> getNext(int num, Num cur) { string nextStringNum = toString(num); if (nextStringNum.size() > cur.size()) { return mp(0, Num(nextStringNum)); } else if (nextStringNum.size() == cur.size()) { // nastLen = 0, by def as nextStringNum.size < 10, or 10 if num = 10^9 if (nextStringNum > cur.pref) { return mp(0, Num(nextStringNum)); } else { return mp(1, addZeros(nextStringNum, 1)); } } else { int nextNumLen = nextStringNum.size(); int curLen = cur.size(); int dod = curLen - nextNumLen; string curPref = cur.pref.substr(0, nextNumLen); if (nextStringNum > curPref) { // only add zeros to make the sizes same return mp(dod, addZeros(nextStringNum, dod)); } else { if (curPref == nextStringNum) { // same prefix, check if we can if (canIncreaseSuffix(cur)) { return mp(dod, Num(cur.pref, cur.nastLen, cur.suffix + 1)); } else { // can increase the rest of the suffix of the prefix of the existing last number string curPrefSuffix = cur.pref.substr(nextNumLen, cur.pref.size() - nextNumLen); if (canIncreaseString(curPrefSuffix)) { int lastSuf = stoi(curPrefSuffix) + 1 + pow(10, curPrefSuffix.size()); string lastSufStringPlus1e9 = toString(lastSuf); return mp(dod, Num(nextStringNum + lastSufStringPlus1e9.substr(1, string::npos), cur.nastLen, 0)); } else { // just add 0 return mp(dod + 1, addZeros(nextStringNum, dod + 1)); } } } else { return mp(dod + 1, addZeros(nextStringNum, dod + 1)); } } } } int main() { scanf("%d", &n); REP(i, n) { scanf("%d", &tab[i]); } Num cur(toString(tab[0])); ll result = 0; FOR(i, 1, n - 1) { pair<int, Num> res = getNext(tab[i], cur); result += res.ST; cur = res.ND; // cout << cur.pref << ' ' << cur.nastLen << ' ' << cur.suffix << endl; } cout << result << endl; 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 | #include <cstdio> #include <iostream> #include <algorithm> #include <cstring> #include <vector> #include <set> #include <map> #include <cmath> #include <list> #include <ctime> #include <sstream> #include <queue> #include <stack> #include <bitset> #include <unordered_set> #include <unordered_map> using namespace std; typedef vector<int> vi; typedef pair<int,int> pii; typedef long long ll; typedef short int sint; #define FOR(x, b, e) for(int x=(b); x<=(e); ++x) #define FORD(x, b, e) for(int x=((int)(b))-1; x>=(e); --x) #define REP(x, n) for(int x=0; x<(n); ++x) #define ALL(c) c.begin(),c.end() #define SIZE(x) ((int)((x).size())) #define PB push_back #define ST first #define ND second #define mp(x,y) make_pair(x,y) #define DEBUG 1 #define debug(x) {if (DEBUG)cerr <<#x <<" = " <<x <<endl; } #define debugv(x) {if (DEBUG) {cerr <<#x <<" = "; FOREACH(it, (x)) cerr <<*it <<", "; cout <<endl; }} #define REMAX(a,b) (a)=max((a),(b)); #define REMIN(a,b) (a)=min((a),(b)); #define wez(n) int (n); scanf("%d",&(n)); #define wez2(n,m) int (n),(m); scanf("%d %d",&(n),&(m)); const int N = 202010; int tab[N]; int n; string toString(int x) { ostringstream s; s << x; return s.str(); } struct Num { string pref; // first 10 chars int nastLen; // length after pref, if pref has 10 chars int suffix; // the ending numbers of the suffix, helpful to check if we can do suffix+1 != 10^suffLen Num(string pref_, int suffLen_ = 0, int suffix_ = 0): pref(pref_), nastLen(suffLen_), suffix(suffix_) { } int size() const { return nastLen + pref.size(); } }; bool canIncreaseSuffix(Num a) { return (a.nastLen > 6 || pow(10, a.nastLen) > a.suffix + 1); } bool canIncreaseString(string x) { REP(i, x.size()) { if (x[i] != '9') { return true; } } return false; } Num addZeros(string nextStringNum, int dod) { while (dod > 0 && nextStringNum.size() < 10) { nextStringNum += '0'; --dod; } return Num(nextStringNum, dod, 0); } pair<int, Num> getNext(int num, Num cur) { string nextStringNum = toString(num); if (nextStringNum.size() > cur.size()) { return mp(0, Num(nextStringNum)); } else if (nextStringNum.size() == cur.size()) { // nastLen = 0, by def as nextStringNum.size < 10, or 10 if num = 10^9 if (nextStringNum > cur.pref) { return mp(0, Num(nextStringNum)); } else { return mp(1, addZeros(nextStringNum, 1)); } } else { int nextNumLen = nextStringNum.size(); int curLen = cur.size(); int dod = curLen - nextNumLen; string curPref = cur.pref.substr(0, nextNumLen); if (nextStringNum > curPref) { // only add zeros to make the sizes same return mp(dod, addZeros(nextStringNum, dod)); } else { if (curPref == nextStringNum) { // same prefix, check if we can if (canIncreaseSuffix(cur)) { return mp(dod, Num(cur.pref, cur.nastLen, cur.suffix + 1)); } else { // can increase the rest of the suffix of the prefix of the existing last number string curPrefSuffix = cur.pref.substr(nextNumLen, cur.pref.size() - nextNumLen); if (canIncreaseString(curPrefSuffix)) { int lastSuf = stoi(curPrefSuffix) + 1 + pow(10, curPrefSuffix.size()); string lastSufStringPlus1e9 = toString(lastSuf); return mp(dod, Num(nextStringNum + lastSufStringPlus1e9.substr(1, string::npos), cur.nastLen, 0)); } else { // just add 0 return mp(dod + 1, addZeros(nextStringNum, dod + 1)); } } } else { return mp(dod + 1, addZeros(nextStringNum, dod + 1)); } } } } int main() { scanf("%d", &n); REP(i, n) { scanf("%d", &tab[i]); } Num cur(toString(tab[0])); ll result = 0; FOR(i, 1, n - 1) { pair<int, Num> res = getNext(tab[i], cur); result += res.ST; cur = res.ND; // cout << cur.pref << ' ' << cur.nastLen << ' ' << cur.suffix << endl; } cout << result << endl; return 0; } |