CodeForces 489C Given Length and Sum of Digits... (贪心)

Given Length and Sum of Digits...

题目链接:

http://acm.hust.edu.cn/vjudge/contest/121332#problem/F

Description

You have a positive integer m and a non-negative integer s. Your task is to find the smallest and the largest of the numbers that have length m and sum of digits s. The required numbers should be non-negative integers written in the decimal base without leading zeroes.

Input

The single line of the input contains a pair of integers m, s (1 ≤ m ≤ 100, 0 ≤ s ≤ 900) — the length and the sum of the digits of the required numbers.

Output

In the output print the pair of the required non-negative integer numbers — first the minimum possible number, then — the maximum possible number. If no numbers satisfying conditions required exist, print the pair of numbers "-1 -1" (without the quotes).

Sample Input

Input
2 15
Output
69 96
Input
3 0
Output
-1 -1

题意:

给出m和s,分别找出最大和最小的非负整数,满足数位长度为m,数位和为s;

题解:

直接贪心枚举每一位就可以了;
最小:能放0就放0; (首位不能是0)
最大:能放9就放9;
WA一次:题目特地强调了非负整数;
那么考虑s为0的情况,并不是全为-1 -1;
当m=1 s=0时应输出0 0;

代码:

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#define LL long long
#define eps 1e-8
#define maxn 150
#define inf 0x3f3f3f3f
#define IN freopen("in.txt","r",stdin);
using namespace std;

int m,s;

int main(int argc, char const *argv[])
{
    //IN;

    int n,s;
    while(scanf("%d %d", &n, &s) != EOF)
    {
        if(n*9=1; i--) {
            int flag = 0;
            if(i==n) flag = 1;
            if(9*(i-1) <= cur-flag) {
                printf("%d", cur-9*(i-1));
                for(int j=1; j<=i-1; j++)
                    printf("9");
                break;
            }
            if(flag) {
                printf("1"); cur-=1;
            } else {
                printf("0");
            }
        }

        printf(" ");

        cur = s;
        for(int i=n; i>=1; i--) {
            if(9 >= cur) {
                printf("%d", cur);
                for(int j=1; j<=i-1; j++)
                    printf("0");
                break;
            }

            printf("9"); cur-=9;
        }

        printf("\n");
    }

    return 0;
}

转载于:https://www.cnblogs.com/Sunshine-tcf/p/5697194.html

你可能感兴趣的:(CodeForces 489C Given Length and Sum of Digits... (贪心))