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| 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 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].
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 |
|
4 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 |
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\n",a%m); 100 } 101 else if(n==1) 102 { 103 printf("%d\n",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\n",ask); 116 } 117 } 118 return 0; 119 }
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\n",x); 99 } 100 else if(z==2) 101 { 102 printf("%llu\n",f3); 103 } 104 else if(z==0) 105 { 106 printf("2\n"); 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\n",ak); 113 } 114 } 115 return 0; 116 }
1065 - Number Sequence &&1070 - Algebraic Problem
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原文地址：http://www.cnblogs.com/zzuli2sjy/p/5595179.html
