题目链接:点我
题意:
给定一列数，每个数对应一个变换，变换为原先数列一些位置相加起来的和，问r次变换后的序列是多少.
思路:
矩阵快速幂,将需要变换的位置变为1, 其他的都为0, 然后快速幂跑R次即可.
代码:
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| #include<cstdio>
#include<algorithm>
#include<cstring>
#include<iostream>
#include<cmath>
using namespace std;
const int mod = 1000;
int b[100];
int n,r;
struct mat{
int a[52][52];
mat(){memset(a, 0, sizeof(a));}
mat operator *(const mat q){
mat c;
for(int i = 1; i <= n; ++i)
for(int j = 1; j <= n; ++j)
if(a[i][j])
for(int k = 1; k <= n; ++k){
c.a[i][k] += a[i][j] * q.a[j][k];
if(c.a[i][k] >= mod) c.a[i][k] %= mod;
}return c;
}
};
mat qpow(mat x, int n){
mat ans;
for(int i = 1; i <= 50; ++i)
ans.a[i][i] = 1;
while(n){
if(n&1) ans = ans * x;
x = x * x;
n >>= 1;
}return ans;
}
int main(){
int t;
scanf("%d", &t);
while(t--){
scanf("%d %d", &n, &r);
for(int i = 1; i <= n; ++i)
scanf("%d", &b[i]);
mat ans;
for(int i = 1; i <= n; ++i){
int num;
scanf("%d", &num);
while(num--){
int k;
scanf("%d", &k);
ans.a[i][k+1] = 1;
}
}ans = qpow(ans,r);
for(int i = 1; i <= n;++i){
int sum = 0;
for(int j = 1; j <= n; ++j){
sum += b[j] * ans.a[i][j];
if(sum>= mod) sum %= mod;
}
if(i!=n)
printf("%d ",sum);
else printf("%d\n",sum);
}
}return 0;
}
|