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  1. #include<iostream>
  2. #include<algorithm>
  3. #include<vector>
  4. #include<string>
  5. #include <cmath>
  6. #include <utility> 
  7. #include <unordered_map>
  8. #include <unordered_set>
  9. #include <map>
  10. #include <cstdint>
  11. #include <iomanip> // Required for std::fixed and std::setprecision
  12. #include<limits>
  13. #include <climits>
  14. using namespace std;
  15.  
  16. // #include <ext/pb_ds/assoc_container.hpp>
  17. // #include <ext/pb_ds/tree_policy.hpp> 
  18. // using namespace __gnu_pbds; 
  19.  
  20. // #define ordered_set tree<int, null_type,less<int>, rb_tree_tag,tree_order_statistics_node_update>
  21. #define int long long
  22.  
  23. // #define setbits(x) __builtin_popcountll((int)x)
  24. #define result(ans) cout<<(ans?"YES":"NO")<<endl;
  25. #define debug cout<<"Chal rha hai be "<<endl;
  26. #define vi vector<int>
  27. #define vs vector<str> 
  28. #define vc vector<char>
  29. #define MOD 1000000007
  30. #define vvi vector<vector<int> >
  31. #define pii pair<int,int>
  32. #define vpi vector<pair<int,int> >
  33. #define ff first
  34. #define ss second
  35. #define mii map<int,int>
  36. #define printv(v) for(int i=0;i<v.size();i++) cout<<v[i]<<" \n"[i==v.size()-1];
  37. #define scanv(v) for(auto &i:v) {cin>>i;}
  38. #define pb push_back
  39. // #define scantree(v) for(int i=0;i<v.size()-1;i++){int x,y;cin>>x>>y;x--;y--;v[x].pb(y);v[y].pb(x);}
  40. #define all(v) v.begin(),v.end()
  41. #define loop(a,b,c) for(int i=a;i<b;i+=c) 
  42. #define MAX 10000
  43. #define max3(a,b,c) max(a,max(b,c))
  44. #define max4(a,b,c,d) max(a,max(b,max(c,d)))
  45. #define mp make_pair
  46.  
  47. vector<int> prefixSum(const vector<int>& arr) {
  48.     int n = arr.size();
  49.     vector<int> prefixSums(n);
  50.     if (n == 0) return prefixSums; 
  51.     prefixSums[0] = arr[0];
  52.     for (int i = 1; i < n; i++) {
  53.         prefixSums[i] = prefixSums[i - 1] + arr[i];
  54.     }
  55.     return prefixSums;
  56. }
  57.  
  58.  
  59.  
  60. // instead of log2
  61. int log(long long x)
  62. {
  63.     return 64 - __builtin_clzll(x) - 1;
  64. }
  65. int maxv(vi v){
  66.     int ans=INT_MIN;
  67.     for(int i=0;i<v.size();i++){
  68.         ans=max(ans,v[i]);
  69.     }
  70.     return ans;
  71. }
  72. int minv(vi v){
  73.     int ans=INT_MAX;
  74.     for(int i=0;i<v.size();i++){
  75.         ans=min(ans,v[i]);
  76.     }
  77.     return ans;
  78. }
  79. int fact(int n)
  80. {
  81.     if(n==0) return 1;
  82.     int res = 1;
  83.     for (int i = 2; i <= n; i++)
  84.         res = res * i;
  85.     return res;
  86. }
  87. int ncr(int n, int r){return fact(n) / (fact(r) * fact(n - r));}
  88. int npr(int n, int r){return fact(n) /  fact(n - r);}
  89.  
  90. int isSorted(vector<int>v){
  91.     for(int i=0; i<v.size()-1; i++){
  92.         if (v[i]>v[i+1]) return 0;
  93.     }
  94.     return 1;
  95. }
  96. int gcd(int a, int b) { 
  97.     return b == 0 ? a : gcd(b, a % b); 
  98. }
  99. int isSubstr(string S1, string S2){
  100.     int M = S1.length();
  101.     int N = S2.length();
  102.     for (int i = 0; i <= N - M; i++) {
  103.         int j;
  104.         for (j = 0; j < M; j++) {
  105.             if (S2[i + j] != S1[j]) {
  106.                 break; 
  107.             }
  108.         }
  109.         if (j == M) {
  110.             return i; 
  111.         }
  112.     }
  113.     return -1; 
  114. }
  115. int freqi(vector<int>a,int b,int c){
  116.     int count =0;
  117.     for(int i=c+1; i<a.size();i++){
  118.         if(a[i]==b){count++;}
  119.     }
  120.     if(count){return count;}
  121.     else {return -1;}
  122. }
  123. int freqs(string a,char b){
  124.     int count =0;
  125.     for(int i=0; i<a.length();i++){
  126.         if(a[i]==b){count++;}}
  127.     if(count){return count;}
  128.     else {return -1;}
  129. }
  130. int lcm(int a,int b){
  131.     return (a*b)/gcd(a,b);
  132. }
  133. bool sortcol(const vector<int>& v1, const vector<int>& v2)
  134. {
  135.     return v1[1] < v2[1];
  136. }
  137. bool c(pair<int,int>p1 , pair<int,int>p2){
  138.    if(p1.first == p2.first){
  139.      return p1.second < p2.second;
  140.    }
  141.    return p1.first  > p2.first;
  142. }
  143. bool isPrime(int n) {
  144.     if (n <= 1) return false;          
  145.     if (n <= 3) return true;           
  146.     if (n % 2 == 0 || n % 3 == 0) return false; 
  147.     for (int i = 5; i * i <= n; i += 6) {
  148.         if (n % i == 0 || n % (i + 2) == 0)
  149.             return false;
  150.     }
  151.     return true;
  152. }
  153. pair<int ,int> smallestPrimeGreaterThan(int num){
  154.     int candidate = num;  
  155.     int count = 0;
  156.     while (!isPrime(candidate)){candidate++;count++;}
  157.     pair<int ,int > x;
  158.     x.first = candidate;
  159.     x.second = count;
  160.     return x;}
  161. vector<int> getFactors(int n) {
  162.     vector<int> factors;
  163.  
  164.     // Loop up to the square root of n
  165.     for (int i = 2; i <= sqrt(n); ++i) {
  166.         if (n % i == 0) {
  167.             factors.push_back(i); // Add the factor
  168.             if (i != n / i) {     // Avoid adding the square root twice
  169.                 factors.push_back(n / i);
  170.             }
  171.         }
  172.     }
  173.     // factors.pb(n);
  174.     // Sort factors in ascending order
  175.     // sort(factors.begin(), factors.end());
  176.     return factors;
  177. }
  178.  
  179.  
  180. vector<bool> sieveOfEratosthenes(int n) {
  181.     // Create a boolean vector initialized to true
  182.     vector<bool> is_prime(n + 1, true);
  183.  
  184.     // Mark 0 and 1 as non-prime
  185.     if (n >= 0) is_prime[0] = false;
  186.     if (n >= 1) is_prime[1] = false;
  187.  
  188.     // Start the Sieve of Eratosthenes algorithm
  189.     for (int i = 2; i * i <= n; i++) {
  190.         if (is_prime[i]) {
  191.             for (int j = i * i; j <= n; j += i) {
  192.                 is_prime[j] = false;
  193.             }
  194.         }
  195.     }
  196.  
  197.     return is_prime;
  198. }
  199.  
  200.  
  201. bool vp_desc(const pair<int, int>& p1, const pair<int, int>& p2) {
  202.     return p1.second > p2.second;
  203. }
  204. int findSubstring(const string& mainStr, const string& subStr) {
  205.     size_t pos = mainStr.find(subStr); 
  206.     if (pos != string::npos) {return pos;}
  207.     else{return -1; }
  208. }
  209. int powerof2(int n) {
  210.     return __builtin_ctz(n);
  211. }
  212. vector<pair<int, int> > getFrequency(const vector<int>& nums) {
  213.     unordered_map<int, int> freqMap;
  214.     for (int num : nums) {
  215.         freqMap[num]++;
  216.     }
  217.  
  218.     vector<pair<int, int> > freqVector;
  219.     for (const auto& entry : freqMap) {
  220.         freqVector.emplace_back(entry.first, entry.second);
  221.     }
  222.  
  223.     sort(freqVector.begin(), freqVector.end());
  224.     return freqVector;
  225. }
  226. int findGCD(const vector<int>& nums) {
  227.     int gcd1 = nums[0];
  228.     for (size_t i = 1; i < nums.size(); ++i) {
  229.         gcd1 = gcd(gcd1, nums[i]);
  230.         if (gcd1 == 1) {
  231.             return 1;
  232.         }
  233.     }
  234.     return gcd1;
  235. }
  236. bool compareBySecond(const pair<int, int>& a, const pair<int, int>& b) {
  237.     return a.second < b.second; // Ascending order by second element
  238. }
  239. int fact_modulo(int n, int mod) {
  240.     int result = 1;
  241.     for (int i = 1; i <= n; i++) {
  242.         result = (result * i) % mod;
  243.     }
  244.     return result;
  245. }
  246.  
  247. int sumv(vi v){
  248.     int sum=0;
  249.     for(auto e:v){
  250.         sum+=e;
  251.     }
  252.     return sum;
  253. }
  254. vector<pair<int, int> > PrimeFactorsfreq(int n) {
  255.     vector<pair<int, int> > primeFactors;
  256.     int count = 0;
  257.     while (n % 2 == 0) {
  258.         count++;
  259.         n /= 2;
  260.     }
  261.     if (count > 0 && 2 != n) { 
  262.         primeFactors.emplace_back(2, count);
  263.     }
  264.     for (int i = 3; i * i <= n; i += 2) {
  265.         count = 0;
  266.         while (n % i == 0) {
  267.             count++;
  268.             n /= i;
  269.         }
  270.         if (count > 0 && i != n) { 
  271.             primeFactors.emplace_back(i, count);
  272.         }
  273.     }
  274.     if (n > 2) {
  275.         primeFactors.emplace_back(n, 1);
  276.     }
  277.     return primeFactors;
  278. }
  279. vi dp;
  280. int ans = 0;
  281. void count(string s ,int n){
  282.     // 
  283.     if(n==0) {ans++;return;};
  284.     if(n<0){return;}
  285.     if(s[n-1]!='0') {count(s,n-1);}
  286.     // if(((s[n-3]-'0')*10 + (s[n-2]-'0'))<=26 && n>=3){
  287.     //     count(s,n-3);
  288.     //     // cout<<((s[n-3]-'0')*10 + (s[n-2]-'0'))<<endl;
  289.     // }
  290.     if(((s[n-2]-'0')*10 + (s[n-1]-'0'))<=26 && n>=2 && s[n-2]!='0'){
  291.         count(s,n-2);
  292.         // cout<<((s[n-3]-'0')*10 + (s[n-2]-'0'))<<endl;
  293.     }
  294.  
  295. }
  296.  
  297. void solve(){
  298.     string s;
  299.     cin>>s;
  300.  
  301.     // int ans = 0;
  302.     // if(s.length()==0){cout<<endl;return;}
  303.     count(s,s.length());
  304.     cout<<ans<<endl;
  305. }
  306. int32_t main(){
  307.     // #ifndef ONLINE_JUDGE
  308.     // freopen("input.txt","r",stdin);
  309.     // freopen("output.txt","w",stdout);
  310.     // #endif
  311.     std::ios::sync_with_stdio(false);
  312.     std::cin.tie(nullptr);
  313.     int t=1;
  314.     // cin >> t;
  315.     while (t--){
  316.         solve();
  317.     }
  318.  
  319.  
  320.     return 0;
  321. }
Success #stdin #stdout 0.01s 5284KB
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https://ideone.com/sOJjQ1
problem ACODE, based on SPOJ submission 34068373 (CPP), 2025-01-16 18:09:36
language:
C++ (gcc 8.3)
created:
1 day ago
visibility:
public

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