@zzzc18 2018-02-23T04:56:13.000000Z 字数 2367 阅读 467

# Codeforces Round #451 (Div. 2) B. Proper Nutrition

Codeforces

## B. Proper Nutrition

time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output

### Description

Vasya has n burles. One bottle of Ber-Cola costs a burles and one Bars bar costs b burles. He can buy any non-negative integer number of bottles of Ber-Cola and any non-negative integer number of Bars bars.

Find out if it's possible to buy some amount of bottles of Ber-Cola and Bars bars and spend exactly n burles.

In other words, you should find two non-negative integers x and y such that Vasya can buy x bottles of Ber-Cola and y Bars bars and x·a + y·b = n or tell that it's impossible.

### Input

First line contains single integer n (1 ≤ n ≤ 10 000 000) — amount of money, that Vasya has.

Second line contains single integer a (1 ≤ a ≤ 10 000 000) — cost of one bottle of Ber-Cola.

Third line contains single integer b (1 ≤ b ≤ 10 000 000) — cost of one Bars bar.

### Output

If Vasya can't buy Bars and Ber-Cola in such a way to spend exactly n burles print «NO» (without quotes).

Otherwise in first line print «YES» (without quotes). In second line print two non-negative integers x and y — number of bottles of Ber-Cola and number of Bars bars Vasya should buy in order to spend exactly n burles, i.e. x·a + y·b = n. If there are multiple answers print any of them.

Any of numbers x and y can be equal 0.

## Solution

$10^7$ 直接暴力枚举就过去了

$a,b$ 满足上式的时候，可以使用拓展欧几里得先求出一组满足上式的整数解 $x,y$

#include<cmath>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;typedef long long LL;LL costa,costb,n;LL Extgcd(LL a,LL b,LL &x,LL &y){    if(b==0){        x=1;y=0;        return a;    }    LL re=Extgcd(b,a%b,x,y);    LL t=x;    x=y;y=(t-(a/b)*y);    return re;}int main(){    scanf("%I64d%I64d%I64d",&n,&costa,&costb);    LL x,y;    LL gcd=Extgcd(costa,costb,x,y);    if(n%gcd!=0){        puts("NO");        return 0;    }    LL tmp=n/gcd;    x*=tmp;y*=tmp;    if(x>=0 && y>=0){        puts("YES");        printf("%I64d %I64d\n",x,y);        return 0;    }    if((x<0 && y<0)||(x<0 && y==0)||(x==0 && y<0)){        puts("NO");        return 0;    }    bool jud=LL(ceil(1.0*(-x)/costb))<=LL(floor(1.0*y/costa));    if(jud){        puts("YES");        LL k=LL(ceil(1.0*(-x)/costb));        x+=k*costb;        y-=k*costa;        printf("%I64d %I64d\n",x,y);    }    else puts("NO");    return 0;}  • 私有
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