ccf-集合竞价-201412-3

这道题数据量很小,所以用前缀和直接暴力解决了

首先将浮点数转换为整数;

int_p=(int)(dou_p*100+0.5); 必须加一个eps,否则浮点数运算会有误差

然后从高到低枚举

 #include <bits/stdc++.h>
 using namespace std;
 typedef long long LL;
 const int N=1e6+;
 const int M=;
 struct node {
     bool by;
     int p; LL num;
 };
 node t[M];
 bool flag[M];
 LL sum1[N], sum2[N];
 int main ()
 {
     string str; int i=,p_max=,p_min=0x3f3f3f3f;
     while (cin>>str) {
         i++;
         if (str[]=='c') { int x; cin>>x; flag[x]=; }
         else {
             double p;
             cin>>p>>t[i].num;
             if (str[]=='b') t[i].by=;
             t[i].p=(int)(p*+0.5);
             p_max=max (p_max,t[i].p);
             p_min=min (p_min,t[i].p);
         }
    }
    int n=i;
    for (i=;i<=n;i++)
         if (!flag[i]) {
             if (t[i].by) sum1[t[i].p]+=t[i].num;
             else         sum2[t[i].p]+=t[i].num;
         }
     for (int i=;i<N;i++) {
         sum1[i]+=sum1[i-];
         sum2[i]+=sum2[i-];
     }
     LL sum=; int ans;
     for (int i=p_max;i>=p_min;i--) {
         LL x=min (sum1[N-]-sum1[i-],sum2[i]);
         if (x>sum) { sum=x; ans=i; }
     }
     int a1=ans/,a2=ans%;
     printf("%d.%02d %lld\n",a1,a2,sum);
     return ;
 }