Contest Print Server
Problem Description
Input
In each case,the first line contains 5 integers n,s,x,y,mod (1<=n<=100, 1<=s,x,y,mod<=10007), and n lines of requests follow. The request is like "Team_Name request p pages" (p is integer, 0<p<=10007, the length of "Team_Name" is no longer than
20), means the team "Team_Name" need p pages to print, but for some un-know reason the printer will break down when the printed pages counter reached s(s is generated by the function s=(s*x+y)%mod ) and then the counter will become 0. In the same time the
last request will be reprint from the very begin if it isn't complete yet(The data guaranteed that every request will be completed in some time).
You can get more from the sample.
Output
Please note that you should print an empty line after each case.
Example Input
23 7 5 6 177Team1 request 1 pagesTeam2 request 5 pagesTeam3 request 1 pages3 4 5 6 177Team1 request 1 pagesTeam2 request 5 pagesTeam3 request 1 pages
Example Output
1 pages for Team15 pages for Team21 pages for Team31 pages for Team13 pages for Team25 pages for Team21 pages for Team3
Hint
Author
#include <iostream> #include <algorithm> #include <stdio.h> using namespace std; typedef long long ll; typedef unsigned long long ull; #define maxn 105 int n,s,x,y,mod; struct str { char name[25]; int p; }; str tt[maxn]; int main() { //read; int cas; scanf("%d",&cas); while(cas--) { scanf("%d%d%d%d%d",&n,&s,&x,&y,&mod); char s1[25],s2[25]; for(int i=1; i<=n; ++i) scanf("%s%s%d%s",tt[i].name,s1,&tt[i].p,s2); int now=s; for(int i=1; i<=n; ++i) { if(tt[i].p<=now) { printf("%d pages for %s\n",tt[i].p,tt[i].name); now-=tt[i].p; } else { printf("%d pages for %s\n",now,tt[i].name); s=(x*s+y)%mod; now=s; --i; } } printf("\n"); } return 0; }