POJ 2196Specialized Four-Digit Numbers简单暴力

D - Specialized Four-Digit Numbers Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%lld & %llu Submit  Status  Practice  POJ 2196

Description

Find and list all four-digit numbers in decimal notation that have the property that the sum of its four digits equals the sum of its digits when represented in hexadecimal (base 16) notation and also equals the sum of its digits when represented in duodecimal (base 12) notation.  For example, the number 2991 has the sum of (decimal) digits 2+9+9+1 = 21. Since 2991 = 1*1728 + 8*144 + 9*12 + 3, its duodecimal representation is 1893  ₁₂, and these digits also sum up to 21. But in hexadecimal 2991 is BAF  ₁₆, and 11+10+15 = 36, so 2991 should be rejected by your program.  The next number (2992), however, has digits that sum to 22 in all three representations (including BB0  ₁₆), so 2992 should be on the listed output. (We don't want decimal numbers with fewer than four digits -- excluding leading zeroes -- so that 2992 is the first correct answer.) 

Input

There is no input for this problem

Output

Your output is to be 2992 and all larger four-digit numbers that satisfy the requirements (in strictly increasing order), each on a separate line with no leading or trailing blanks, ending with a new-line character. There are to be no blank lines in the output. The first few lines of the output are shown below.

Sample Input

There is no input for this problem

Sample Output

2992 2993 2994 2995 2996 2997 2998 2999 ...

简单遍历2992->10000

#include <iostream> #include <cstdio> #include <cstdlib> #include <cmath> #include <algorithm> #include <climits> #include <cstring> #include <string> #include <set> #include <map> #include <queue> #include <stack> #include <vector> #include <list> #define rep(i,m,n) for(i=m;i<=n;i++) #define rsp(it,s) for(set<int>::iterator it=s.begin();it!=s.end();it++) const int inf_int = 2e9; const long long inf_ll = 2e18; #define inf_add 0x3f3f3f3f #define mod 1000000007 #define vi vector<int> #define pb push_back #define mp make_pair #define fi first #define se second #define pi acos(-1.0) #define pii pair<int,int> #define Lson L, mid, rt<<1 #define Rson mid+1, R, rt<<1|1 const int maxn=5e2+10; using namespace std; typedef long long ll; typedef unsigned long long ull; inline int read(){int ra,fh;char rx;rx=getchar(),ra=0,fh=1; while((rx<'0'||rx>'9')&&rx!='-')rx=getchar();if(rx=='-') fh=-1,rx=getchar();while(rx>='0'&&rx<='9')ra*=10,ra+=rx-48, rx=getchar();return ra*fh;} //#pragma comment(linker, "/STACK:102400000,102400000") ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);} ll qpow(ll p,ll q){ll f=1;while(q){if(q&1)f=f*p;p=p*p;q>>=1;}return f;} int main() { int t; int sum = 0,sum1,sum2; for(int i=2992;i<10000;i++) { sum = 0; sum1 = 0; sum2=0; t = i; while(t!=0) { sum += t; t/=10; } t= i; for(int j=3;j>=0;j--) { int x = pow(12,j); sum1 += t/x ; t %= x; } t= i; for(int j=3;j>=0;j--) { int x = pow(16,j); sum2 += t/x ; t %= x; } //cout << sum<<" " <<sum1<<" " <<sum2<<endl; if(sum1==sum2&&sum1==sum) { cout << i ; if(i!=9929) cout << endl; } } return 0; }