time limit per test 2 seconds
memory limit per test 256 megabytes
Description
Facer’s pet cat just gave birth to a brood of little cats. Having considered the health of those lovely cats, Facer decides to make the cats to do some exercises. Facer has well designed a set of moves for his cats. He is now asking you to supervise the cats to do his exercises. Facer’s great exercise for cats contains three different moves:
g i : Let the ith cat take a peanut.
e i : Let the ith cat eat all peanuts it have.
s i j : Let the ith cat and jth cat exchange their peanuts.
All the cats perform a sequence of these moves and must repeat it m times! Poor cats! Only Facer can come up with such embarrassing idea.
You have to determine the final number of peanuts each cat have, and directly give them the exact quantity in order to save them.
Input
The input file consists of multiple test cases, ending with three zeroes “0 0 0”. For each test case, three integers n, m and k are given firstly, where n is the number of cats and k is the length of the move sequence. The following k lines describe the sequence.
(m≤1,000,000,000, n≤100, k≤100)
Output
For each test case, output n numbers in a single line, representing the numbers of peanuts the cats have.
Sample Input
3 1 6
g 1
g 2
g 2
s 1 2
g 3
e 2
0 0 0
Sample Output
2 0 1
上一次做矩阵快速幂专题大概是半年前了…..昨天在子扬学长的提醒下决定重开这个专题（假）复习（真预习）。
这不做不知道….一做我还真不会….看到这题的一瞬间我大脑的第一反应是—线段树！然而…..这不是区间更新根本没必要用线段树(≡ω≡．)
这题我是wa了七发才过，最后发现是题目里面EOF的要求是ending with three zeroes “0 0 0”. 然后我写的是n&&m&&k，这样m其实是可以为0的但是我这里break掉了 ╮(╯_╰)╭，改成&&（m+n+k)就好了或者直接声明break就好了……以后写题不能为了便利取一些“巧”啊啊啊啊啊。
下面是代码，裸裸的矩阵快速幂。
这里用n+1行单位矩阵表示n个单位的val（peanut）变化，不多解释了，线性代数的知识，m轮操作直接矩阵连乘就好。
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
using namespace std;
typedef long long ll;
ll n,m,k;
struct Matrix
{
ll mi[150][150];
};
Matrix a;
Matrix Mul(Matrix a,Matrix b)
{
Matrix c;
memset(c.mi,0,sizeof(c.mi));
for(int i=1;i<=n;i++)
{
for(int j=1;j<=n;j++)
{
if(a.mi[i][j]==0)
continue;
else
{
for(int k=1;k<=n;k++)
{
if(b.mi[j][k]==0)
continue;
c.mi[i][k]+=a.mi[i][j]*b.mi[j][k];
}
}
}
}
return c;
}
Matrix qic_pow(ll m)
{
Matrix unit;
memset(unit.mi,0,sizeof(unit.mi));
for(int i=1;i<=n;i++)
unit.mi[i][i]=1;
while(m)
{
if(m&1)
unit=Mul(unit,a);
a=Mul(a,a);
m/=2;
}
return unit;
}
int main()
{
while(scanf("%lld%lld%lld",&n,&m,&k)!=EOF&&(m+n+k))
{
n++;
memset(a.mi,0,sizeof(a.mi));
for(int i=1;i<=n;i++)
a.mi[i][i]=1;
char s[5];
while(k--)
{
scanf("%s",s);
if(s[0]=='g')
{
ll x;
scanf("%lld",&x);
a.mi[x][n]++;
}
if(s[0]=='e')
{
ll x;
scanf("%lld",&x);
for(int i=1;i<=n;i++)
a.mi[x][i]=0;
}
if(s[0]=='s')
{
ll x,y;
scanf("%lld%lld",&x,&y);
ll temp;
for(int i=1;i<=n;i++)
{
swap(a.mi[x][i],a.mi[y][i]);
}
}
}
Matrix ans=qic_pow(m);
for(int i=1;i<n;i++)
{
if(i!=n-1)
printf("%lld ",ans.mi[i][n]);
else
printf("%lld\n",ans.mi[i][n]);
}
}
}