zoukankan      html  css  js  c++  java
  • poj-2142-exgcd/解的和最小

    Time Limit: 5000MS   Memory Limit: 65536K
    Total Submissions: 8816   Accepted: 3833

    Description

    Ms. Iyo Kiffa-Australis has a balance and only two kinds of weights to measure a dose of medicine. For example, to measure 200mg of aspirin using 300mg weights and 700mg weights, she can put one 700mg weight on the side of the medicine and three 300mg weights on the opposite side (Figure 1). Although she could put four 300mg weights on the medicine side and two 700mg weights on the other (Figure 2), she would not choose this solution because it is less convenient to use more weights. 
    You are asked to help her by calculating how many weights are required. 

    Input

    The input is a sequence of datasets. A dataset is a line containing three positive integers a, b, and d separated by a space. The following relations hold: a != b, a <= 10000, b <= 10000, and d <= 50000. You may assume that it is possible to measure d mg using a combination of a mg and b mg weights. In other words, you need not consider "no solution" cases. 
    The end of the input is indicated by a line containing three zeros separated by a space. It is not a dataset.

    Output

    The output should be composed of lines, each corresponding to an input dataset (a, b, d). An output line should contain two nonnegative integers x and y separated by a space. They should satisfy the following three conditions. 
    • You can measure dmg using x many amg weights and y many bmg weights. 
    • The total number of weights (x + y) is the smallest among those pairs of nonnegative integers satisfying the previous condition. 
    • The total mass of weights (ax + by) is the smallest among those pairs of nonnegative integers satisfying the previous two conditions.

    No extra characters (e.g. extra spaces) should appear in the output.

    Sample Input

    700 300 200
    500 200 300
    500 200 500
    275 110 330
    275 110 385
    648 375 4002
    3 1 10000
    0 0 0

    Sample Output

    1 3
    1 1
    1 0
    0 3
    1 1
    49 74
    3333 1

    Source

        

      给出两种砝码质量为a,b,问能不能测出质量c的东西,求最小的砝码数量,如果有多个方案考虑总质量最小的方案。

      a*x+b*y=c ,求满足方程的 abs(x)+abs(y)的最小值。做两次exgcd取一个最优的。一次让x为最小正整数,一次让y为最小正整数。

    (但我总觉得应该让求出来的解在减去一个d'/d看会不会更优,这里没进行这一步但是A了。

     1 #include<iostream>
     2 #include<cstdio>
     3 #include<cmath>
     4 using namespace std;
     5 #define LL long long 
     6 #define mp make_pair
     7 #define pb push_back
     8 #define inf 0x3f3f3f3f
     9 void exgcd(LL a,LL b,LL &d,LL &x,LL &y){
    10     if(!b){d=a;x=1;y=0;}
    11     else{exgcd(b,a%b,d,y,x);y-=(a/b)*x;}
    12 }
    13 int main(){
    14     LL a,b,c,d,x1,x2,y1,y2;
    15     while(cin>>a>>b>>c&&(a||b||c)){
    16         exgcd(a,b,d,x1,y1);
    17         exgcd(b,a,d,x2,y2);
    18         if(c%d){
    19             puts("no solution");
    20         }
    21         else{
    22             LL d1=b/d,d2=a/d;
    23             x1=x1*c/d,y1=y1*c/d;
    24             x2=x2*c/d,y2=y2*c/d;
    25             x1=(x1%d1+d1)%d1,y1=(c-a*x1)/b;
    26             x2=(x2%d2+d2)%d2,y2=(c-b*x2)/a;
    27             x1=fabs(x1),y1=fabs(y1);
    28             x2=fabs(x2),y2=fabs(y2);
    29             if(x1+y1<x2+y2) cout<<x1<<' '<<y1<<endl;
    30             else if(x1+y1>x2+y2) cout<<y2<<' '<<x2<<endl;
    31             else{
    32                 if(x1*a+y1*b<x2*a+y2*b) cout<<x1<<' '<<y1<<endl;
    33                 else cout<<y2<<' '<<x2<<endl;
    34             }
    35         }
    36     }
    37     return 0;
    38 }
  • 相关阅读:
    flask项目--认证方案Json Web Token(JWT)
    分布式设计-集群
    分布式设计-哨兵
    分布式设计--数据库主从
    leetcode 221 Maximal Square
    LeetCode222 Count Complete Tree Nodes
    在windows 、linux下读取目录下所有文件名
    leetcode 229 Majority Element II
    leetcode 233 Number of Digit One
    leetcode 238 Product of Array Except Self
  • 原文地址:https://www.cnblogs.com/zzqc/p/9478782.html
Copyright © 2011-2022 走看看