Description
Input
第1行:10個空格分開的整數: N, a, b, c, d, e, f, g, h, M
Output
第1行:滿足總重量最輕,且用度之和最大的N頭奶牛的總體重模M後的餘數。
Sample Input
Sample Output
HINT
樣例說明:公式生成的體重和有用度分別爲: 體重:5, 6, 9, 14, 21, 30 有用度:0, 1, 8, 27, 64, 125.
模擬就好了……
#include<cstdio>
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<cmath>
#include<queue>
#include<deque>
#include<set>
#include<map>
#include<ctime>
#define LL long long
#define inf 0x7ffffff
#define pa pair<int int>
using namespace std;
inline LL read()
{
LL x=0,f=1;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
return x*f;
}
LL n,a,b,c,d,e,f,g,h,m;
struct cow{
LL w,u;
}aaa[1500010];
inline bool cmp(const cow &a,const cow &b)
{
return a.u>b.u||a.u==b.u&&a.w<b.w;
}
inline void work()
{
for(int i=0;i<3*n;i++)
{
LL t1=i%d;
LL t2=(t1*t1)%d;
LL t5=(((t2*t2)%d)*t1)%d;
aaa[i].w=(a*t5+b*t2+c)%d;
LL s1=i%h;
LL s3=(((s1*s1)%h)*s1)%h;
LL s5=(((s3*s1)%h)*s1)%h;
aaa[i].u=(e*s5+f*s3+g)%h;
}
}
int main()
{
n=read();
a=read();
b=read();
c=read();
d=read();
e=read();
f=read();
g=read();
h=read();
m=read();
work();
sort(aaa,aaa+3*n,cmp);
LL sum=0;
for (int i=0;i<n;i++) sum=(sum+aaa[i].w)%m;
printf("%lld\n",sum);
}