Problem Description
Tom and Jerry are playing a game with tubes and pearls. The rule of the game is:
- Tom and Jerry come up together with a number K.
- Tom provides N tubes. Within each tube, there are several pearls. The number of pearls in each tube is at least 1 and at most N.
- Jerry puts some more pearls into each tube. The number of pearls put into each tube has to be either 0 or a positive multiple of K. After that Jerry organizes these tubes in the order that the first tube has exact one pearl, the 2nd tube has exact 2 pearls, …, the Nth tube has exact N pearls.
- If Jerry succeeds, he wins the game, otherwise Tom wins.
Write a program to determine who wins the game according to a given N, K and initial number of pearls in each tube. If Tom wins the game, output “Tom”, otherwise, output “Jerry”.
Input
The first line contains an integer M (M<=500), then M games follow. For each game, the first line contains 2 integers, N and K (1 <= N <= 100, 1 <= K <= N), and the second line contains N integers presenting the number of pearls in each tube.
Output
For each game, output a line containing either “Tom” or “Jerry”.
Sample Input
2
5 1
1 2 3 4 5
6 2
1 2 3 4 5 5
Sample Output
Jerry
Tom
Source
2014上海全国邀请赛——题目重现（感谢上海大学提供题目）
题意：
初始状态每个tube里有一定的珍珠，
现在只能往tube中添加 a*k 或 0 个珍珠；
其中 k 给定， a ∈ Z；
问最后排列后可不可以使得数目为;
1,2,3…n；
我们只需将每个 tube 的数量的可能情况都列出来
（前提 <=n ）；
然后进行一次二分图匹配即可；
看最后可不可以完全匹配；
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<bitset>
#include<ctime>
#include<deque>
#include<stack>
#include<functional>
#include<sstream>
#include<cctype>
//#pragma GCC optimize("O3")
using namespace std;
#define maxn 200005
#define inf 0x3f3f3f3f
#define INF 0x7fffffff
#define rdint(x) scanf("%d",&x)
#define rdllt(x) scanf("%lld",&x)
typedef long long ll;
typedef unsigned long long ull;
typedef unsigned int U;
#define ms(x) memset((x),0,sizeof(x))
const int mod = 1e9 + 7;
#define Mod 20100403
#define sq(x) (x)*(x)
#define eps 1e-10
const int N = 1505;
inline int rd() {
int x = 0;
char c = getchar();
bool f = false;
while (!isdigit(c)) {
if (c == '-') f = true;
c = getchar();
}
while (isdigit(c)) {
x = (x << 1) + (x << 3) + (c ^ 48);
c = getchar();
}
return f ? -x : x;
}
ll gcd(ll a, ll b) {
return b == 0 ? a : gcd(b, a%b);
}
ll sqr(ll x) { return x * x; }
int T;
int n, k;
int vis[2000];
int match[1000];
int G[1000][1000];
bool dfs(int x) {
for (int i = 1; i <= n; i++) {
if (G[x][i] && !vis[i]) {
vis[i] = 1;
if (!match[i] || dfs(match[i])) {
match[i] = x; return true;
}
}
}
return false;
}
int main()
{
//ios::sync_with_stdio(false);
rdint(T);
while (T--) {
rdint(n); rdint(k);
ms(vis); ms(match); ms(G);
for (int i = 1; i <= n; i++) {
int x; rdint(x);
while (x <= n) {
G[x][i] = 1; x += k;
}
}
int full = 0;
for (int i = 1; i <= n; i++) {
ms(vis);
full += dfs(i);
}
if (full == n)cout << "Jerry" << endl;
else cout << "Tom" << endl;
}
return 0;
}
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<bitset>
#include<ctime>
#include<deque>
#include<stack>
#include<functional>
#include<sstream>
#include<cctype>
//#pragma GCC optimize("O3")
using namespace std;
#define maxn 200005
#define inf 0x3f3f3f3f
#define INF 0x7fffffff
#define rdint(x) scanf("%d",&x)
#define rdllt(x) scanf("%lld",&x)
typedef long long ll;
typedef unsigned long long ull;
typedef unsigned int U;
#define ms(x) memset((x),0,sizeof(x))
const int mod = 1e9 + 7;
#define Mod 20100403
#define sq(x) (x)*(x)
#define eps 1e-10
const int N = 1505;
inline int rd() {
int x = 0;
char c = getchar();
bool f = false;
while (!isdigit(c)) {
if (c == '-') f = true;
c = getchar();
}
while (isdigit(c)) {
x = (x << 1) + (x << 3) + (c ^ 48);
c = getchar();
}
return f ? -x : x;
}
ll gcd(ll a, ll b) {
return b == 0 ? a : gcd(b, a%b);
}
ll sqr(ll x) { return x * x; }
int T;
int n, k;
int vis[2000];
int match[1000];
int G[1000][1000];
bool dfs(int x) {
for (int i = 1; i <= n; i++) {
if (G[x][i] && !vis[i]) {
vis[i] = 1;
if (!match[i] || dfs(match[i])) {
match[i] = x; return true;
}
}
}
return false;
}
int main()
{
//ios::sync_with_stdio(false);
rdint(T);
while (T--) {
rdint(n); rdint(k);
ms(vis); ms(match); ms(G);
for (int i = 1; i <= n; i++) {
int x; rdint(x);
while (x <= n) {
G[x][i] = 1; x += k;
}
}
int full = 0;
for (int i = 1; i <= n; i++) {
ms(vis);
full += dfs(i);
}
if (full == n)cout << "Jerry" << endl;
else cout << "Tom" << endl;
}
return 0;
}
