【枚舉】XVII Open Cup named after E.V. Pankratiev Stage 4: Grand Prix of SPb, Sunday, Octorber 9, 2016 Problem D. Cutting Potatoes
阿新 • • 發佈:2017-09-28
題意 clas tag ble cpp rand ring ++i break
題意:有n個土豆,每個有體積V(i),你可以將每個土豆等分為不超過K份,問你最大塊和最小塊比值最小為多少。
直接枚舉切法,只有n*K種,然後保證其為最大塊,去算其他塊的切法,即讓其他塊切得盡可能大即可。O(n*n*K)。
#include<cstdio> #include<cstring> using namespace std; int n,K,a[105],ans1=150,ans2=1,x[105],ansx[105]; int main(){ // freopen("d.in","r",stdin); scanf("%d%d",&n,&K); for(int i=1;i<=n;++i){ scanf("%d",&a[i]); } for(int i=1;i<=n;++i){ for(int j=1;j<=K;++j){ int tmp1=150,tmp2=1; x[i]=j; bool flag=1; for(int k=1;k<=n;++k){ if(k!=i){ x[k]=a[k]*j/a[i]; if(a[k]*j%a[i]!=0){ ++x[k]; } if(x[k]>K){ flag=0; break; } if(a[k]*tmp2<x[k]*tmp1){ tmp1=a[k]; tmp2=x[k]; } } } if(flag && a[i]*tmp2*ans2<ans1*j*tmp1){ ans1=a[i]*tmp2; ans2=j*tmp1; memcpy(ansx,x,sizeof(x)); } } } for(int i=1;i<n;++i){ printf("%d ",ansx[i]); } printf("%d\n",ansx[n]); return 0; }
【枚舉】XVII Open Cup named after E.V. Pankratiev Stage 4: Grand Prix of SPb, Sunday, Octorber 9, 2016 Problem D. Cutting Potatoes