Time
Limit: 3000/1000 MS (Java/Others) Memory Limit:
65535/32768 K (Java/Others)
Total Submission(s):
3940 Accepted Submission(s):
1023
1 #include<stdio.h> 2 #include<string.h> 3 #define maxn 10002 4 #define modc 1000000007 5 int circle[maxn]; 6 int main() 7 { 8 int test,n,t,i; 9 __int64 k; 10 scanf("%d",&test); 11 while(test--) 12 { 13 scanf("%d%d%I64d",&n,&t,&k); 14 for(i=1; i<=n ;i++) 15 scanf("%d",circle+i); 16 int te=t; 17 __int64 res=1; 18 while(te) //用一个快速幂求解 19 { 20 if(te&1) 21 { 22 res=((res%modc)*(k%modc))%modc; 23 te--; 24 continue; 25 } 26 k*=k; 27 k%=modc; 28 te>>=1; 29 } 30 t%=n; //去掉重复的周期 31 int pos; 32 i=1; 33 if(i-t<=0)pos=i-t+n; 34 else pos=i-t; 35 printf("%d",((circle[pos]%modc)*(res%modc)%modc)); 36 for(i=2;i<=n;i++) 37 { 38 if(i-t<=0)pos=i-t+n; 39 else pos=i-t; 40 printf(" %d",((circle[pos]%modc)*(res%modc)%modc)); 41 } 42 putchar(10); 43 } 44 return 0; 45 }
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原文:http://www.cnblogs.com/gongxijun/p/3607907.html