Description
Tom and Jerry are playing a game with tubes and pearls. The rule of the game is:
1) Tom and Jerry come up together with a number K.
2) 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.
3) 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.
4) 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”.
/*
Tom和Jerry做游戏，给定N个管子，每个管子上面有一定数目的珍珠。
现在 Jerry 先开始在管子上面再放一些珍珠
放上的珍珠数必须是K的倍数，或者是 0
最后将管子排序，如果可以做到第 i 个管子上面有 i 个珍珠，
则Jerry胜出，反之Tom胜出
解题思路：
贪心算法
将个管子按管子中的珍珠数量升序排序，则满足条件的时候第i个管子应该有i个珍珠。
每个管子开始遍历，如果第i个管子有i个珍珠，则进行下一次循环；
如果大于i个珍珠，则说明不满足条件
如果小于i个珍珠，则加上k个珍珠，重新按升序排序
（这里一定要重新排序，当你加上 k 之后，a[i] 有可能是 i 之后的某一个个管子里的珍珠数
例如：5 2 1 2 3 3 4
当 i=4 时，a[i]=3, a[i]=a[i]+k=5, 排序之后 ，5 正好是 i=5 时 的珍珠数，而 i=5 时珍珠数 4 恰好是 a[4] 的珍珠数 ）
*/
代码：
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<iostream>
#include<algorithm>
using namespace std;
int main()
{
int m,n,k,i,b,flag=0,sum=0;
int a[101];
cin>>m;
while(m--)
{
flag=0;
sum=0;
memset(a,0,sizeof(a));
cin>>n>>k;
for(i=0;i<=n-1;i++)
scanf("%d",&a[i]);
sort(a,a+n);
for(i=0;i<=n-1;i++){
if(a[i]==i+1){
sum++;
continue;
}
while(a[i]<i+1){
a[i]+=k;
sort(a+i,a+n);
}
if(a[i]>i+1){
flag=1;
break;
}
}
if(sum==n){
printf("Jerry\n");
continue;
}
if(flag==1)
printf("Tom\n");
else
printf("Jerry\n");
}
}/*
我的错误思路 po 出来长个教训
1 处：
这样是错误的，当 a[i] < k 时，b 为 两者的差，如果直接判断 差是不是 k 的倍数，太过果断了，因为很有可能 ａ［ｉ］＋ｋ 和以后某一项的珍珠数互换就能实现 JERRY 赢（例如正确代码中的例子）
*/
错误的代码：
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<iostream>
#include<algorithm>
using namespace std;
int main()
{
int m,n,k,i,b,flag=0,sum=0;
int a[101];
cin>>m;
while(m--)
{
flag=0;
sum=0;
memset(a,0,sizeof(a));
cin>>n>>k;
for(i=0;i<=n-1;i++)
scanf("%d",&a[i]);
sort(a,a+n);
for(i=0;i<=n-1;i++){
if(a[i]==i+1){
sum++;
continue;
}
if(a[i]<i+1){
b=(i+1)-a[i];
// 1.
if(b%k==0)
flag=1;
else{
flag=0;
break;
}
}
if(a[i]>i+1){
flag=0;
break;
}
}
if(sum==n||flag==1)
printf("Jerry\n");
else
printf("Tom\n");
}
}