UVa 1583 Digit Generator

Description

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For a positive integer N , the digit-sum of N is defined as the sum of N itself and its digits. When M is the digitsum of N , we call N a generator of M .

For example, the digit-sum of 245 is 256 (= 245 + 2 + 4 + 5). Therefore, 245 is a generator of 256.

Not surprisingly, some numbers do not have any generators and some numbers have more than one generator. For example, the generators of 216 are 198 and 207.

You are to write a program to find the smallest generator of the given integer.

Input

Your program is to read from standard input. The input consists of T test cases. The number of test cases T is given in the first line of the input. Each test case takes one line containing an integer N , 1≤ \leN≤ \le100, 000 .

Output

Your program is to write to standard output. Print exactly one line for each test case. The line is to contain a generator of N for each test case. If N has multiple generators, print the smallest. If N does not have any generators, print 0.

The following shows sample input and output for three test cases.

Sample Input

3

216

121

2005

Sample Output

198

0

1979

题意就是求求一个最小的数字x，是的x加上x的各个位置上的和是y。完全可以暴力做，但是有一点要注意，如果直接从0枚举到这个数的话肯定会超时的，但是有一点可以肯定的是，这些书做多不超过1e7，也就是说，各个位置上的数字和最多不会超过70，所以只需要枚举max（0，y-70）到y-1就行。

AC代码：

[code]//
//  Created by  CQUWEL
//  Copyright (c) 2015年 CQUWEL. All rights reserved.
//
#include <iostream>
#include <cstring>
#include <cstdio>
#include <cstdlib>
#include <cctype>
#include <cmath>
#include <string>
#include <vector>
#include <list>
#include <map>
#include <queue>
#include <stack>
#include <set>
#include <algorithm>
#include <numeric>
#include <functional>
#define cir(i,a,b)  for (int i=a;i<=b;i++)
#define CIR(j,a,b)  for (int j=a;j>=b;j--)
#define CLR(x) memset(x,0,sizeof(x))
#define root 1,n,1
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define filetest freopen("C:\\Users\\john\\Desktop\\in.txt","r",stdin)
#define BUG() printf("In here\n")
#define seevalue(x) printf("x=%d\n",x)
#define INF 0x3f3f3f3f
const double eps=1e-9;
typedef long long  ll;
using namespace std;

string s;

ll digsum(ll a){
    int sum=0;
    ll x=a;
    while (x){
        sum+=x%10;
        x/=10;
    }
    return sum+a;
}

int main(){
    int n;
    cin >> n;
    for (int i=1;i<=n;i++){
        ll a;
        cin >> a;
        int flag=0;

        for (ll i=max((ll)0,a-70);i<a;i++){
            if (digsum(i)==a) {
                printf("%lld\n",i);
                flag=1;
                break;
            }
        }
        if (!flag) printf("0\n");
    }
    return 0;
}