BZOJ1407: [Noi2002]Savage

1407: [Noi2002]Savage

Time Limit: 5 Sec  Memory Limit: 64 MB
Submit: 2357  Solved: 1056
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Description

Input

第1行为一个整数N(1<=N<=15)，即野人的数目。

第2行到第N+1每行为三个整数Ci, Pi, Li表示每个野人所住的初始洞穴编号，每年走过的洞穴数及寿命值。

(1<=Ci,Pi<=100, 0<=Li<=10^6 )

Output

仅包含一个数M，即最少可能的山洞数。输入数据保证有解，且M不大于10^6。

Sample Input

3
1 3 4
2 7 3
3 2 1

Sample Output

6
//该样例对应于题目描述中的例子。

HINT

Source

 

【题解】

题目中说答案不超过1e6，暗示我们枚举答案。设有m个洞穴，如何判断是否可行？

对于任意i,j(i≠j0,满足对任意x，有Ci + xPi = Cj + xPj (mod m)无解或x的最小正整数解>min(l[i],l[j])

枚举i,j，判断即可

这里前者到求最小正整数解的方法

对于方程ax + by = c

设g = gcd(a,b),g | c

设a' = a/g, b' = b/g, c' = c/g

转化为a'x + b'y = c'

用exgcd求a'x + b'y = 1的解

求得x为x0,于是a'x + b'y = c'中x解集为

x = x0*c' + kb'/gcd(a', b') = x0 + kb',k∈Z

若b'为负数，需要把b'变成正数，这样变可行因为只影响y不影响x

xmin = (x * c' % b' + b') % b'

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#include <queue>
#include <vector>
#define min(a, b) ((a) < (b) ? (a) : (b))
#define max(a, b) ((a) > (b) ? (a) : (b))
#define abs(a) ((a) < 0 ? (-1 * (a)) : (a))
inline void swap(int &a, int &b)
{
    int tmp = a;a = b;b = tmp;
}
inline void read(int &x)
{
    x = 0;char ch = getchar(), c = ch;
    while(ch < '0' || ch > '9') c = ch, ch = getchar();
    while(ch <= '9' && ch >= '0') x = x * 10 + ch - '0', ch = getchar();
    if(c == '-') x = -x;
}

const int INF = 0x3f3f3f3f;
const int MAXN = 15;

int c[MAXN + 5], p[MAXN + 5], l[MAXN + 5], n, ma; 

void exgcd(int a, int b, int &x, int &y)
{
    if(!b){x = 1;y = 0;return;}
    exgcd(b, a % b, y, x);
    y -= a / b * x;
    return;
}
int gcd(int a, int b)
{
    return !b ? a : gcd(b, a % b);
}

bool judge(int m)
{
    for(register int i = 1;i <= n;++ i) for(register int j = i + 1;j <= n;++j)
    {
        //先求方程 (pj - pi)x + my = ci - cj的最小正整数解x
        int P = p[j] - p[i], C = c[i] - c[j], g = gcd(P, m), M = m, x = 0, y = 0;
        if(C % g != 0) continue;
        P /= g, C /= g, M /= g;
        exgcd(P, M, x, y);
        M = M < 0 ? -M : M;
        x = (x * C % M + M) % M;
        if(!x) x += M;
        if(x <= min(l[i], l[j])) return 0;
    }
    return 1;
}

int main()
{
    read(n);
    for(register int i = 1;i <= n;++ i) read(c[i]), read(p[i]), read(l[i]), ma = max(ma, c[i]);
    for(register int i = ma;;++ i)
        if(judge(i)){printf("%d", i);return 0;}
} 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#include <queue>
#include <vector>
#define min(a, b) ((a) < (b) ? (a) : (b))
#define max(a, b) ((a) > (b) ? (a) : (b))
#define abs(a) ((a) < 0 ? (-1 * (a)) : (a))
inline void swap(int &a, int &b)
{
    int tmp = a;a = b;b = tmp;
}
inline void read(int &x)
{
    x = 0;char ch = getchar(), c = ch;
    while(ch < '0' || ch > '9') c = ch, ch = getchar();
    while(ch <= '9' && ch >= '0') x = x * 10 + ch - '0', ch = getchar();
    if(c == '-') x = -x;
}

const int INF = 0x3f3f3f3f;
const int MAXN = 15;

int c[MAXN + 5], p[MAXN + 5], l[MAXN + 5], n, ma; 

void exgcd(int a, int b, int &x, int &y)
{
    if(!b){x = 1;y = 0;return;}
    exgcd(b, a % b, y, x);
    y -= a / b * x;
    return;
}
int gcd(int a, int b)
{
    return !b ? a : gcd(b, a % b);
}

bool judge(int m)
{
    for(register int i = 1;i <= n;++ i) for(register int j = i + 1;j <= n;++j)
    {
        //先求方程 (pj - pi)x + my = ci - cj的最小正整数解x
        int P = p[j] - p[i], C = c[i] - c[j], g = gcd(P, m), M = m, x = 0, y = 0;
        if(C % g != 0) continue;
        P /= g, C /= g, M /= g;
        exgcd(P, M, x, y);
        M = M < 0 ? -M : M;
        x = (x * C % M + M) % M;
        if(!x) x += M;
        if(x <= min(l[i], l[j])) return 0;
    }
    return 1;
}

int main()
{
    read(n);
    for(register int i = 1;i <= n;++ i) read(c[i]), read(p[i]), read(l[i]), ma = max(ma, c[i]);
    for(register int i = ma;;++ i)
        if(judge(i)){printf("%d", i);return 0;}
}