1065

1065 - Number Sequence

Time Limit: 2 second(s)

Memory Limit: 32 MB

Let's define another number sequence, given by the following function:

f(0) = a

f(1) = b

f(n) = f(n-1) + f(n-2), n > 1

When a = 0 and b = 1, this sequence gives the Fibonacci sequence. Changing the values of a and b, you can get many different sequences. Given the values of a, b, you have to find the last m digits of f(n).

Input

Input starts with an integer T (≤ 10000), denoting the number of test cases.

Each test case consists of a single line containing four integers a b n m. The values of a and b range in [0,100], value of n ranges in [0, 10⁹] and value of m ranges in [1, 4].

Output

For each case, print the case number and the last m digits of f(n). However, do NOT print any leading zero.

Sample Input

Output for Sample Input

0 1 11 3

0 1 42 4

0 1 22 4

0 1 21 4

Case 1: 89

Case 2: 4296

Case 3: 7711

Case 4: 946

Special Thanks: Jane Alam Jan (Solution, Dataset)

矩阵快速幂水题

  1 #include<stdio.h>
  2 #include<algorithm>
  3 #include<iostream>
  4 #include<string.h>
  5 #include<stdlib.h>
  6 #include<queue>
  7 #include<math.h>
  8 #include<vector>
  9 using namespace std;
 10 typedef  long long LL;
 11 typedef struct pp
 12 {
 13         LL m[4][4];
 14         pp()
 15         {
 16                 memset(m,0,sizeof(m));
 17         }
 18 } maxtr;
 19 maxtr E()
 20 {
 21         maxtr ans;
 22         int i,j;
 23         for(i=0; i<4; i++)
 24         {
 25                 for(j=0; j<4; j++)
 26                 {
 27                         if(i==j)
 28                         {
 29                                 ans.m[i][j]=1;
 30                         }
 31                         else ans.m[i][j]=0;
 32                 }
 33         }
 34         return ans;
 35 }
 36 void Init (maxtr *p)
 37 {   int i,j;
 38         for(i=0; i<2; i++)
 39         {
 40                 for(j=0; j<2; j++)
 41                 {
 42                         if(i==1&&j==1)
 43                         {
 44                                 p->m[i][j]=0;
 45                         }
 46                         else  p->m[i][j]=1;
 47                 }
 48         }
 49 }
 50 maxtr quick(maxtr ans ,int m,int mod)
 51 {
 52         maxtr ak=E();
 53         int i,j;
 54         while(m)
 55         {
 56                 if(m&1)
 57                 {
 58                         maxtr cc;
 59                         for(i=0; i<2; i++)
 60                         {
 61                                 for(j=0; j<2; j++)
 62                                 {
 63                                         for(int s=0; s<2; s++)
 64                                         {
 65                                                 cc.m[i][j]=(cc.m[i][j]+ans.m[i][s]*ak.m[s][j]%mod)%mod;
 66                                         }
 67                                 }
 68                         }
 69                         ak=cc;
 70                 }
 71                 maxtr cc;
 72                 for(i=0; i<2; i++)
 73                 {
 74                         for(j=0; j<2; j++)
 75                         {
 76                                 for(int s=0; s<2; s++)
 77                                 {
 78                                         cc.m[i][j]=(cc.m[i][j]+ans.m[i][s]*ans.m[s][j]%mod)%mod;
 79                                 }
 80                         }
 81                 }
 82                 ans=cc;
 83                 m/=2;
 84         }
 85         return ak;
 86 }
 87 int main(void)
 88 {
 89         int i,j,k;
 90         int s;
 91         scanf("%d",&k);
 92         int a,b,n,m;
 93         for(s=1; s<=k; s++)
 94         {
 95                 scanf("%d %d %d %d",&a,&b,&n,&m);
 96                 printf("Case %d: ",s);
 97                 if(n==0)
 98                 {
 99                         printf("%d
",a%m);
100                 }
101                 else if(n==1)
102                 {
103                         printf("%d
",b%m);
104                 }
105                 else
106                 {
107                         maxtr ak;
108                         Init(&ak);
109                         int mod=1;
110                         for(i=1;i<=m;i++)
111                             mod*=10;
112                         ak=quick(ak,n-1,mod);
113                         LL ask=ak.m[0][0]*b+ak.m[0][1]*a;
114                         ask%=mod;
115                         printf("%lld
",ask);
116                 }
117         }
118         return 0;
119 }

1065

  1 #include<stdio.h>
  2 #include<algorithm>
  3 #include<iostream>
  4 #include<string.h>
  5 #include<stdlib.h>
  6 #include<queue>
  7 #include<math.h>
  8 #include<vector>
  9 using namespace std;
 10 typedef unsigned long long LL;
 11 typedef long long L;
 12 typedef struct pp
 13 {
 14         LL m[4][4];
 15         pp()
 16         {
 17                 memset(m,0,sizeof(m));
 18         }
 19 } maxtr;
 20 maxtr E()
 21 {
 22         int i,j;
 23         maxtr ans;
 24         for(i=0; i<=3; i++)
 25         {
 26                 for(j=0; j<=3; j++)
 27                 {
 28                         if(i==j)
 29                                 ans.m[i][j]=1;
 30                         else ans.m[i][j]=0;
 31                 }
 32         }
 33         return ans;
 34 }
 35 void Init(maxtr *p,LL x,LL y)
 36 {
 37         int i,j;
 38         memset(p->m,0,sizeof(p->m));
 39         p->m[0][0]=x;
 40         p->m[0][1]=-y;
 41         p->m[1][0]=1;
 42 }
 43 maxtr quick(maxtr ak,LL m)
 44 {
 45         int i,j,s;
 46         maxtr ac=E();
 47 
 48         while(m)
 49         {
 50                 if(m&1)
 51                 {
 52                         maxtr cc;
 53                         for(i=0; i<2; i++)
 54                         {
 55                                 for(j=0; j<2; j++)
 56                                 {
 57                                         for(s=0; s<2; s++)
 58                                         {
 59                                                 cc.m[i][j]=(cc.m[i][j]+ak.m[i][s]*ac.m[s][j]);
 60                                         }
 61                                 }
 62                         }
 63                         ac=cc;
 64                 }
 65                 maxtr cc;
 66                 for(i=0; i<2; i++)
 67                 {
 68                         for(j=0; j<2; j++)
 69                         {
 70                                 for(s=0; s<2; s++)
 71                                 {
 72                                         cc.m[i][j]=(cc.m[i][j]+ak.m[i][s]*ak.m[s][j]);
 73                                 }
 74                         }
 75                 }
 76                 ak=cc;
 77                 m/=2;
 78         }
 79         return ac;
 80 }
 81 int main(void)
 82 {
 83         int i,j,k;
 84         scanf("%d",&k);
 85         int s;
 86         for(s=1; s<=k; s++)
 87         {
 88                 LL x,y,z;
 89                 scanf("%llu %llu %llu",&x,&y,&z);
 90                 LL f1=1;
 91                 LL f2=x;
 92                 LL f3=x*x-2*y;
 93                 maxtr ans;
 94                 Init(&ans,x,y);
 95                 printf("Case %d: ",s);
 96                 if(z==1)
 97                 {
 98                         printf("%llu
",x);
 99                 }
100                 else if(z==2)
101                 {
102                         printf("%llu
",f3);
103                 }
104                 else if(z==0)
105                 {
106                     printf("2
");
107                 }
108                 else
109                 {
110                        ans= quick(ans,z-2);
111                         LL ak=ans.m[0][0]*f3+ans.m[0][1]*f2;
112                         printf("%llu
",ak);
113                 }
114         }
115         return 0;
116 }

1070

油！油！you@

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原文地址：https://www.cnblogs.com/zzuli2sjy/p/5595179.html