水题,就是对每组的$max(\frac{at_1(100-k)}{100.0}+b{t_2},\frac{bt_1(100-k)}{100.0}+a{t_2})$排序,如果相等,再让序号排1
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using namespace std;
struct prog{
int k;
double high;
}ans[1005];
bool cmp(prog a,prog b)
{
if(a.high==b.high)
return a.k<b.k;
return a.high>b.high;
}
int main()
{
int n,t1,t2,k;
while(~scanf("%d%d%d%d",&n,&t1,&t2,&k))
{
int i;
for(i=0;i<n;i++)
{
int a,b;
scanf("%d%d",&a,&b);
ans[i].k=i;
ans[i].high=max(a*t1*(100-k)/100.0+b*t2*1.0,b*t1*(100-k)/100.0+a*t2*1.0);
}
sort(ans,ans+n,cmp);
for(i=0;i<n;i++)
printf("%d %.2lf\n",ans[i].k+1,ans[i].high);
}
return 0;
}