這道題數據量很小,所以用前綴和直接暴力解決了

首先將浮點數轉換為整數;

int_p=(int)(dou_p*100+0.5); 必須加一個eps,否則浮點數運算會有誤差

然後從高到低枚舉

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 typedef long long LL;
 4 const int N=1e6+7;
 5 const int M=5007;
 6 struct node {
 7     bool by;
 8     int p; LL num;
 9 };
10 node t[M]; 
 
11 bool flag[M];
12 LL sum1[N], sum2[N]; 
13 int main ()
14 {
15     string str; int i=0,p_max=0,p_min=0x3f3f3f3f;
16     while (cin>>str) {
17         i++;
18         if (str[0]==‘c‘) { int x; cin>>x; flag[x]=1; }
19         else {
20             double p;
21             cin>>p>>t[i].num; 
 
22             if (str[0]==‘b‘) t[i].by=1;
23             t[i].p=(int)(p*100+0.5);
24             p_max=max (p_max,t[i].p);
25             p_min=min (p_min,t[i].p);
26         }
27    }
28    int n=i;
29    for (i=1;i<=n;i++) 
30         if (!flag[i]) {
31             if (t[i].by) sum1[t[i].p]+=t[i].num;
 
32             else         sum2[t[i].p]+=t[i].num;
33         }
34     for (int i=1;i<N;i++) {
35         sum1[i]+=sum1[i-1];
36         sum2[i]+=sum2[i-1];
37     }
38     LL sum=0; int ans;
39     for (int i=p_max;i>=p_min;i--) {
40         LL x=min (sum1[N-1]-sum1[i-1],sum2[i]);
41         if (x>sum) { sum=x; ans=i; }
42     }
43     int a1=ans/100,a2=ans%100;
44     printf("%d.%02d %lld\n",a1,a2,sum);
45     return 0;
46 }

ccf-集合競價-201412-3