UVA 10306 e-Coins(二维完全背包）
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At the Department for Bills and Coins,an extension of today's monetary system has newly been proposed,in order to make it fit the new economy better. A number of new so called e-coins will be produced,which,in addition to having a value in the normal sense of today,also have an InfoTechnological value. The goal of this reform is,of course,to make justice to the economy of numerous dotcom companies which,despite the fact that they are low on money surely have a lot of IT inside. All money of the old kind will keep its conventional value and get zero InfoTechnological value. To successfully make value comparisons in the new system,something called the e-modulus is introduced. This is calculated asSQRT(X*X+Y*Y),where X and Y hold the sums of the conventional and InfoTechnological values respectively. For instance,money with a conventional value of $3 altogether and an InfoTechnological value of $4 will get an e-modulus of $5. Bear in mind that you have to calculate the sums of the conventional and InfoTechnological values separately before you calculate the e-modulus of the money. To simplify the move to e-currency,you are assigned to write a program that,given the e-modulus that shall be reached and a list of the different types of e-coins that are available,calculates the smallest amount of e-coins that are needed to exactly match the e-modulus. There is no limit on how many e-coins of each type that may be used to match the given e-modulus. Input A line with the number of problems n (0<n<=100),followed by n times:
When more than one number is present on a line,they will be separated by a space. Between each problem,there will be one blank line.
Output The output consists of n lines. Each line contains either a single integer holding the number of coins necessary to reach the specified e-modulus S or,if S cannot be reached,the string "not possible". Sample Input: 3 3 20 3 5 Sample Output: not possible
2维背包：题意：两个属性x,y,求最少的数使(x1+x2+.xk)^2+(y1+y2+..yk)^2=s*s； 斟酌背包:dp[i][j]表示属性为i,j的时候数。 #include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
typedef long long LL;
using namespace std;
#define REPF( i,a,b ) for ( int i = a ; i <= b ; ++ i )
#define REP( i,n ) for ( int i = 0 ; i < n ; ++ i )
#define CLEAR( a,x ) memset ( a,x,sizeof a )
const int INF=0x3f3f3f3f;
const int maxn=330;
int dp[maxn][maxn];
int w1[55],w2[55];
int t,n,s;
int main()
{
std::ios::sync_with_stdio(false);
cin>>t;
while(t--)
{
cin>>n>>s;
CLEAR(dp,INF);
dp[0][0]=0;
REP(i,n)
cin>>w1[i]>>w2[i];
REP(i,n)
{
REPF(j,w1[i],s)
{
REPF(k,w2[i],s)
{
if(dp[j-w1[i]][k-w2[i]]!=INF)
dp[j][k]=min(dp[j][k],dp[j-w1[i]][k-w2[i]]+1);
}
}
}
int ans=INF;
REPF(i,s)
{
REPF(j,s)
{
if(dp[i][j]!=INF&&i*i+j*j==s*s)
ans=min(ans,dp[i][j]);
}
}
if(ans!=INF) cout<<ans<<endl;
else cout<<"not possible"<<endl;
}
return 0;
}
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