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Codeforces Round #359 (Div. 2) C. Robbers' watch Violent Enumeration

2022-08-09 07:04:00 【swust_fang】

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The meaning of the question is really annoying. At the end, I know that n m is actually the number of digits in the two time zones of the limited table, so when the number is not enough to fill the time zone, the front is automatically filled with zeros

Thinking: First of all, if there cannot be repeated numbers, the total number of digits in the hours and minutes must not be greater than 7.

7 to the 7th power is only 823543

So we can enumerate i from 0 to n, and j from 0 to m.Why is the complexity correct?

In fact, it is to consider the maximum number of LenN+LenM<=7 and n*m (Len indicates how many digits)

Consider two limits N1 bits, M6 bits, the maximum 7*7^6 is just the 7th power of 7

N4 bits, M3 bits, maximum 7^3*7^4 or 7^7, less than 1e6 complexity

//#pragma comment (linker, "/STACK:102400000,102400000")#include#include#include#include#include#include#include#include#include#include#include#include#include#include#define myself i,l,r#define lson i<<1#define rson i<<1|1#define Lson i<<1,l,mid#define Rson i<<1|1,mid+1,r#define half (l+r)/2#define lowbit(x) x&(-x)#define min4(a, b, c, d) min(min(a,b),min(c,d))#define min3(x, y, z) min(min(x,y),z)#define max3(x, y, z) max(max(x,y),z)#define max4(a, b, c, d) max(max(a,b),max(c,d))//freopen("E://1.in","r",stdin);//freopen("E://1.out","w",stdout);typedef unsigned long long ull;typedef long long ll;#define pii make_pair#define pr pairconst int inff = 0x3f3f3f3f;const int dir[4][2] = {0, 1, 0, -1, 1, 0, -1, 0};const int mdir[8][2] = {0, 1, 0, -1, 1, 0, -1, 0, 1, 1, -1, 1, 1, -1, -1, -1};const double eps = 1e-10;const double PI = acos(-1.0);const double E = 2.718281828459;using namespace std;const long long inFF = 9223372036854775807;const int mod=1e9+7;const int maxn=1e5+5;int a[7];int len1,len2;int cal(ll x){x--;if(x==0) return 1;int ans=0;while(x){x/=7;ans++;}return ans;}bool check(int x,int y){int cnt=0;memset(a,0,sizeof(a));while(x){a[x%7]++;x/=7;cnt++;}a[0]+=len1-cnt;cnt=0;while(y){a[y%7]++;y/=7;cnt++;}a[0]+=len2-cnt;for(int i=0;i<=6;i++)if(a[i]>=2)return 0;return 1;}int main(){ll n,m;cin>>n>>m;len1=cal(n),len2=cal(m);if(len1+len2>7){cout<<0<

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