2008-2009 ACM-ICPC, NEERC, Moscow Subregional Contest B Bankrupt Broker

题目连接：

http://codeforces.com/gym/100861/attachments

题解：

比较水的题目，基本看的懂题意就能打。
这里需要一下离散化。

代码：

#include <cmath>#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>using namespace std;#define met(a,b) memset(a,b,sizeof(a))#define inf 0x3f3f3f3fconst int maxn = 7e4+10;struct node1{    int cnt;    int logo;}num1[maxn];struct node2{    int price;    int day;}num2[maxn];int change1[maxn],len1;int change2[maxn],len2;int ans[maxn];int cmp1(node1 a,node1 b){    return a.cnt>b.cnt;}int cmp2(node2 a,node2 b){    return a.price>b.price;}int id(int num){    return lower_bound(change2,change2+len2,num)-change2+1;}void init(){    for(int i=0;i<maxn;i++)    {        num1[i].cnt=0;        num1[i].logo=-1;    }    for(int i=0;i<maxn;i++)        num2[i].price=num2[i].day=0;    met(change1,0),met(change2,0);    len1=0,len2=0;    met(ans,0);}int main(){    init();    int F,A,B,C;    scanf("%d%d%d%d",&F,&A,&B,&C);    int n,m;    scanf("%d%d",&n,&m);    for(int i=0;i<n;i++)        for(int j=0;j<m;j++)        {            int x;            scanf("%d",&x);            change1[len1++]=x;            change2[len2++]=x;        }    sort(change2,change2+len2);    len2=unique(change2,change2+len2)-change2;    int len3=0;    for(int i=0;i<len1;i++)    {        if(num1[id(change1[i])].logo==-1)        {            len3++;            num1[(id(change1[i]))].logo=change1[i];        }        num1[(id(change1[i]))].cnt++;    }    sort(num1+1,num1+maxn,cmp1);    num2[1].price=F;    num2[1].day=1;    for(int i=2;i<=len3;i++)    {        num2[i].price=((A*num2[i-1].price+B)%C)+1;        num2[i].day=i;    }    sort(num2+1,num2+len3+1,cmp2);    for(int i=1;i<=len3;i++)        ans[num2[i].day]=num1[i].logo;    printf("%d\n",len3);    for(int i=1;i<=len3;i++)    {        if(i==1)            printf("%d",ans[i]);        else            printf(" %d",ans[i]);    }    printf("\n");}