MUTC8 J-The More The Better
The More The Better
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2568 Accepted Submission(s): 668
Problem Description
Given an sequence of numbers {X
1, X
2, ... , X
n}, where X
k = (A * k + B) % mod. Your task is to find the maximum sub sequence {Y
1, Y
2, ... , Y
m} where every pair of (Y
i, Y
j) satisfies Y
i + Y
j <= L (1 ≤ i < j ≤ m),
and every Yi <= L (1 ≤ i ≤ m ).
Now given n, L, A, B and mod, your task is to figure out the maximum m described above.
Now given n, L, A, B and mod, your task is to figure out the maximum m described above.
Input
Multiple test cases, process to the end of input. Every test case has a single line. A line of 5 integers: n, L, A, B and mod. (1 ≤ n ≤ 2*10
7, 1 ≤ L ≤ 2*10
9, 1 ≤ A, B, mod ≤ 10
9)
Output
For each case, output m in one line.
Sample Input
1 8 2 3 6 5 8 2 3 6
Sample Output
1 4
Source
2012 Multi-University Training Contest 8
Recommend
zhuyuanchen520
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简单题，首先想到所有小于 L/2 的，统统可以放进来，最后，按照题意，还可能可以放一个大于 L/2 的数进来，当小于 L/2 的数里面的最大值加上这个大于 L/2 的数的和小于 L 时，答案加一。最后要注意所有数都小于 L/2 的处理。O(n) 算法可过此题。
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/** head-file **/
#include <iostream>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <string>
#include <vector>
#include <queue>
#include <stack>
#include <list>
#include <set>
#include <map>
#include <algorithm>
/** define-for **/
#define REP(i, n) for (int i=0;i<int(n);++i)
#define FOR(i, a, b) for (int i=int(a);i<int(b);++i)
#define DWN(i, b, a) for (int i=int(b-1);i>=int(a);--i)
#define REP_1(i, n) for (int i=1;i<=int(n);++i)
#define FOR_1(i, a, b) for (int i=int(a);i<=int(b);++i)
#define DWN_1(i, b, a) for (int i=int(b);i>=int(a);--i)
#define REP_N(i, n) for (i=0;i<int(n);++i)
#define FOR_N(i, a, b) for (i=int(a);i<int(b);++i)
#define DWN_N(i, b, a) for (i=int(b-1);i>=int(a);--i)
#define REP_1_N(i, n) for (i=1;i<=int(n);++i)
#define FOR_1_N(i, a, b) for (i=int(a);i<=int(b);++i)
#define DWN_1_N(i, b, a) for (i=int(b);i>=int(a);--i)
/** define-useful **/
#define clr(x,a) memset(x,a,sizeof(x))
#define sz(x) int(x.size())
#define see(x) cerr<<#x<<" "<<x<<endl
#define se(x) cerr<<" "<<x
#define pb push_back
#define mp make_pair
/** test **/
#define Display(A, n, m) { \
REP(i, n){ \
REP(j, m) cout << A[i][j] << " "; \
cout << endl; \
} \
}
#define Display_1(A, n, m) { \
REP_1(i, n){ \
REP_1(j, m) cout << A[i][j] << " "; \
cout << endl; \
} \
}
using namespace std;
/** typedef **/
typedef long long LL;
/** Add - On **/
const int direct4[4][2]={ {0,1},{1,0},{0,-1},{-1,0} };
const int direct8[8][2]={ {0,1},{1,0},{0,-1},{-1,0},{1,1},{1,-1},{-1,1},{-1,-1} };
const int direct3[6][3]={ {1,0,0},{0,1,0},{0,0,1},{-1,0,0},{0,-1,0},{0,0,-1} };
const int MOD = 1000000007;
const int INF = 0x3f3f3f3f;
const long long INFF = 1LL << 60;
const double EPS = 1e-9;
const double OO = 1e15;
const double PI = acos(-1.0); //M_PI;
const int maxn=11111111;
LL ai;
LL n,L,A,B,mod;
LL mx,mi;
LL ans;
int main()
{
while (cin>>n>>L>>A>>B>>mod)
{
mx=0;
mi=L;
ans=0;
ai=B;
REP_1(i,n)
{
ai+=A;
while (ai>=mod) ai-=mod;
if (ai<=L/2)
{
ans++;
mx=max(mx,ai);
}
else
{
mi=min(mi,ai);
}
}
if (mi+mx<=L) ans++;
cout<<ans<<endl;
}
return 0;
}