Gym 100531D Digits (暴力)

题意：给定一个数字，问你找 n 个数，使得这 n 个数各位数字之和都相等，并且和最小。

析：暴力，去枚举和是 1 2 3...，然后去选择最小的。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#define debug puts("+++++")
//#include <tr1/unordered_map>
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;
//using namespace std :: tr1;

typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const LL LNF = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 50000 + 5;
const LL mod = 1e3 + 7;
const int N = 1e6 + 5;
const int dr[] = {-1, 0, 1, 0, 1, 1, -1, -1};
const int dc[] = {0, 1, 0, -1, 1, -1, 1, -1};
const char *Hex[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
inline LL gcd(LL a, LL b){  return b == 0 ? a : gcd(b, a%b); }
inline int gcd(int a, int b){  return b == 0 ? a : gcd(b, a%b); }
inline int lcm(int a, int b){  return a * b / gcd(a, b); }
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline int Min(int a, int b){ return a < b ? a : b; }
inline int Max(int a, int b){ return a > b ? a : b; }
inline LL Min(LL a, LL b){ return a < b ? a : b; }
inline LL Max(LL a, LL b){ return a > b ? a : b; }
inline bool is_in(int r, int c){
    return r >= 0 && r < n && c >= 0 && c < m;
}
int a[25], b[25];

bool judge(int x, int y){
    int tmp = 0;
    while(x){
        tmp += x % 10;
        x /= 10;
    }
    return tmp == y;
}

int main(){
    freopen("digits.in", "r", stdin);
    freopen("digits.out", "w", stdout);
    while(scanf("%d", &n) == 1){
        memset(a, 0, sizeof a);
        memset(b, 0, sizeof b);
        for(int i = 1; i < 100000; ++i)
            for(int j = 1; j <= 20; ++j)
                if(a[j] < n && judge(i, j)){ b[j] += i; ++a[j]; }
        int ans = INF;
        for(int i = 1; i <= 20; ++i)  if(a[i] == n){
            ans = Min(ans, b[i]);
        }
        cout << ans << endl;
    }
    return 0;
}