[关闭]
@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

考试的时候看错题,以致于耽误太长时间,我把数据范围看成 了。
直接暴力枚举就过去了

对于更大数据的 算法如下:
题目要求是否有一组非负整数对 使得 成立

首先,根据裴蜀定理,如果 显然不成立
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
此时,可以通过调整 的值使得 均非负,那么由下式,可以进行调整:

这样,我们得到

进而

所以,如果满足

就说明有一组可行非负整数解
至于将这个解构造出来,直接将任意一个满足条件的 代入即可

这相当于求不定方程非负整数解的通法吧,之前一直不知道怎么搞


  1. #include<cmath>
  2. #include<cstdio>
  3. #include<cstring>
  4. #include<algorithm>
  5. using namespace std;
  6. typedef long long LL;
  7. LL costa,costb,n;
  8. LL Extgcd(LL a,LL b,LL &x,LL &y){
  9. if(b==0){
  10. x=1;y=0;
  11. return a;
  12. }
  13. LL re=Extgcd(b,a%b,x,y);
  14. LL t=x;
  15. x=y;y=(t-(a/b)*y);
  16. return re;
  17. }
  18. int main(){
  19. scanf("%I64d%I64d%I64d",&n,&costa,&costb);
  20. LL x,y;
  21. LL gcd=Extgcd(costa,costb,x,y);
  22. if(n%gcd!=0){
  23. puts("NO");
  24. return 0;
  25. }
  26. LL tmp=n/gcd;
  27. x*=tmp;y*=tmp;
  28. if(x>=0 && y>=0){
  29. puts("YES");
  30. printf("%I64d %I64d\n",x,y);
  31. return 0;
  32. }
  33. if((x<0 && y<0)||(x<0 && y==0)||(x==0 && y<0)){
  34. puts("NO");
  35. return 0;
  36. }
  37. bool jud=LL(ceil(1.0*(-x)/costb))<=LL(floor(1.0*y/costa));
  38. if(jud){
  39. puts("YES");
  40. LL k=LL(ceil(1.0*(-x)/costb));
  41. x+=k*costb;
  42. y-=k*costa;
  43. printf("%I64d %I64d\n",x,y);
  44. }
  45. else puts("NO");
  46. return 0;
  47. }
添加新批注
在作者公开此批注前,只有你和作者可见。
回复批注