hdu 5610 /BC 69A Baby Ming and Weight lifting
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首先如果c不能整除2 则无法满足要求
然后把c分成两部分 看哑铃单侧重量c/2的情况
暴力枚举a、b中较大者的个数即可 看剩余需求重量能否整除较小数
#include<iostream>#include<cstdio>#include<cstring>#include<cctype>#include<cmath>#include<vector>#include<queue>#include<map>#include<algorithm>#include<set>#define scnaf scanf#define cahr char#define bug puts("bugbugbug");using namespace std;typedef long long ll;const int mod=1000000007;const int maxn=50+5;const int inf=1e9;const int maxe=200000;int main(){ int T_T,a,b,c; scanf("%d",&T_T); while(T_T--){ scanf("%d%d%d",&a,&b,&c); if(c%2){ puts("Impossible"); continue; } c/=2; int ans=inf,flag=0; if(a>b) { swap(a,b); flag=1; } for(int i=c/b;i>=0;i--) if((c-i*b)%a==0){ ans=i; break; } if(ans==inf) puts("Impossible"); else{ if(!flag) printf("%d %d\n",2*(c-ans*b)/a,2*ans); else printf("%d %d\n",2*ans,2*(c-ans*b)/a); } } return 0;}
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