Codeforces Round #513 by Barcelona bootcamp (rated, div. 1 + div. 2)

Codeforces Round #513 by Barcelona bootcamp (rated, div. 1 + div. 2)

• Solved:2 out of 8 ...
• rank:2730 Unrated
• A. Phone Numbers
• Difficulty: Universal group.
• Simulation can be:
• Summary: Do not waste too much time on simple questions, to improve the accuracy of the problem.

  #include<cstdio>#include<cstring>#include<algorithm>#include<iostream>using namespace std;const int N = 100 + 10;int cnt =0;int main(){    int n; scanf("%d", &n);    for(int i=1;i<=n;i++) {        char ch;        cin>>ch;        if(ch == ‘8‘) cnt++;    }    int ans=0;    while(n){        if(cnt&&n>=11) cnt--,n-=11,ans++;        else break;    }    printf("%d",ans);    return 0;}

• B. Maximum Sum of Digits
• Difficulty: Increase group D1T1?
• greedy
  • for numbers less than 10, a = N, b = 0
  • for numbers greater than 10, we choose to use as many ⑨ as possible to gather a until a is greater than N, and then the highest bit of a is the highest bit of N-1, which constructs a. 。
  • and then B = n-a.
  • Tips:
    • pow () function or something ... or write yourself a long long fast power bar ...
    • Sample Touching ... Fortunately played a watch to find the law ...
  • Summary:
    • conclusion or greedy problem must be a table to write to check the correctness of the conclusion.
    • to prevent intermediate bursts of long long, in the problem of big data, you can open long long as far as possible.

  #include <cstdio> #include <cstring> #include <algorithm> #include <cmath>using    namespace Std;long long cal (Long long a) {long long ans=0;        while (a) {long long dig=a%10;        Ans+=dig;    a/=10; } return ans;    Long Long Fpow (long long A,long long B) {long long ans=1;        for (; b;b>>=1) {if (b&1) ans*=a;    A*=a; } return ans;    int main () {long long n; scanf ("%i64d", &n);    Long long ans;    Long Long A;    if (n<10) a= N;        else {a = 0;long long cnt=0;        while (a<=n) a=a*10+9,cnt++;        if (a>n) a/=10,cnt--;        Long long Top;long long x=n;            while (x) {if (x/10>0) TOP=X/10;        x/=10;        } top--;        Long Long Pw=fpow (10,cnt);    A= (long Long) top*pw+a;    } long long B;    if (n<10) b=0;    else {b = n-a;    } ans = cal (a) + cal (b);    printf ("%i64d", ans); return 0;}  

Codeforces Round #513 by Barcelona bootcamp (rated, div. 1 + div. 2)