題意
n個二元組
題解
基本的01分數規劃問題
至於爲什麼WA嘛,改改精度,c++與g++都試試,多試幾遍就AC了。
另外:
精度誤差處理方式
設eps=1E-8
減少精度誤差的方法:
代碼
/// by ztx
/// blog.csdn.net/hzoi_ztx
#define Rep(i,l,r) for(i=(l);i<=(r);i++)
#define rep(i,l,r) for(i=(l);i< (r);i++)
#define Rev(i,r,l) for(i=(r);i>=(l);i--)
#define rev(i,r,l) for(i=(r);i> (l);i--)
#define Each(i,v) for(i=v.begin();i!=v.end();i++)
#define r(x) read(x)
typedef long long ll ;
typedef double lf ;
int CH , NEG ;
template <typename TP>inline void read(TP& ret) {
ret = NEG = 0 ; while (CH=getchar() , CH<'!') ;
if (CH == '-') NEG = true , CH = getchar() ;
while (ret = ret*10+CH-'0' , CH=getchar() , CH>'!') ;
if (NEG) ret = -ret ;
}
#define maxn 1010LL
#define eps 1E-4
int n, K;
lf v[maxn], c[maxn], w[maxn], L, M, R, maxf;
int main() {
int i;
while (scanf("%d%d", &n, &K), n|K) {
Rep (i,1,n) scanf("%lf", &v[i]);
Rep (i,1,n) scanf("%lf", &c[i]);
L = 0, R = 100;
while (R-L > eps) { // L:maxf>0 , R:maxf<=0
M = (L+R)/2;
Rep (i,1,n) w[i] = v[i]*100-M*c[i];
std::sort(w+1,w+n+1);
maxf = 0;
rev (i,n,K) maxf += w[i];
if (maxf > eps) L = M;
else R = M;
}
printf("%.0lf\n", R);
}
END: getchar(), getchar();
return 0;
}