bzoj2018 [Usaco2009 Nov]农场技艺大赛

[Usaco2009 Nov]农场技艺大赛

Time Limit: 10 Sec Memory Limit: 162 MB
Submit: 311 Solved: 153
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Description

Input

第1行：10个空格分开的整数: N, a, b, c, d, e, f, g, h, M

Output

第1行：满足总重量最轻，且用度之和最大的N头奶牛的总体重模M后的余数。

Sample Input

2 0 1 5 55555555 0 1 0 55555555 55555555

Sample Output

HINT

样例说明：公式生成的体重和有用度分别为：体重：5, 6, 9, 14, 21, 30 有用度：0, 1, 8, 27, 64, 125.

Source

Silver

题解：模拟

 1 #include<cstring>
 2 #include<cmath>
 3 #include<iostream>
 4 #include<algorithm>
 5 #include<cstdio>
 6 #include<cstdlib>
 7 
 8 #define ll long long
 9 using namespace std;
10 inline int read()
11 {
12     int x=0,f=1;char ch=getchar();
13     while(ch<'0'||ch>'9'){if (ch=='-') f=-1;ch=getchar();}
14     while(ch>='0'&&ch<='9'){x=(x<<3)+(x<<1)+ch-'0';ch=getchar();}
15     return x*f;
16 }
17 ll n,a,b,c,d,e,f,g,h,m;  
18 struct cow
19 {  
20     ll w,u;  
21 }aaa[1500010];  
22 inline bool cmp(cow a,cow b)  
23 {  
24     return a.u>b.u||a.u==b.u&&a.w<b.w;  
25 }  
26 inline void work()  
27 {  
28     for(int i=0;i<3*n;i++)  
29     {  
30         ll t1=i%d;  
31         ll t2=(t1*t1)%d;  
32         ll t5=(((t2*t2)%d)*t1)%d;  
33         aaa[i].w=(a*t5+b*t2+c)%d;  
34         ll s1=i%h;  
35         ll s3=(((s1*s1)%h)*s1)%h;  
36         ll s5=(((s3*s1)%h)*s1)%h;  
37         aaa[i].u=(e*s5+f*s3+g)%h;  
38     }  
39 }  
40 int main()  
41 {  
42     n=read();  
43     a=read();  
44     b=read();  
45     c=read();  
46     d=read();  
47     e=read();  
48     f=read();  
49     g=read();  
50     h=read();  
51     m=read();  
52     work();  
53     sort(aaa,aaa+3*n,cmp);  
54     ll sum=0;  
55     for (int i=0;i<n;i++) sum=(sum+aaa[i].w)%m;  
56     printf("%lld
",sum);  
57 }