华中农业大学第五届程序设计大赛网络同步赛-D
Problem D: GCD
Time Limit: 1 Sec Memory Limit: 1280 MBSubmit: 179 Solved: 25
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Description
XiaoMingfoundthecomputetimeof
Input
The first line is an positive integer T. (
Output
In each test case, output the compute result of
Sample Input
1
1 2 3
Sample Output
1
HINT
1 #include <cstdio> 2 #include <cstring> 3 #include <iostream> 4 #include <algorithm> 5 #define LL long long 6 #define MAXN 100 7 8 using namespace std; 9 10 int MOD; 11 12 struct Matrix 13 { 14 LL a[MAXN][MAXN]; 15 int r, c; 16 }; 17 18 Matrix ori, res; 19 20 void init() 21 { 22 memset(res.a, 0, sizeof(res.a)); 23 res.r = 2; res.c = 2; 24 for(int i = 1; i <= 2; i++) 25 res.a[i][i] = 1; 26 ori.r = 2; ori.c = 2; 27 ori.a[1][1] = ori.a[1][2] = ori.a[2][1] = 1; 28 ori.a[2][2] = 0; 29 } 30 31 Matrix multi(Matrix x, Matrix y) 32 { 33 Matrix z; 34 memset(z.a, 0, sizeof(z.a)); 35 z.r = x.r, z.c = y.c; 36 for(int i = 1; i <= x.r; i++) 37 { 38 for(int k = 1; k <= x.c; k++) 39 { 40 if(x.a[i][k] == 0) continue; 41 for(int j = 1; j<= y.c; j++) 42 z.a[i][j] = (z.a[i][j] + (x.a[i][k] * y.a[k][j]) % MOD) % MOD; 43 } 44 } 45 return z; 46 } 47 void Matrix_mod(int n) 48 { 49 while(n) 50 { 51 if(n & 1) 52 res = multi(ori, res); 53 ori = multi(ori, ori); 54 n >>= 1; 55 } 56 printf("%lld\n", res.a[1][2] % MOD); 57 } 58 59 int main() 60 { 61 int T, n, m; 62 scanf("%d", &T); 63 while(T--) 64 { 65 scanf("%d%d%d", &n, &m, &MOD); 66 LL pos = __gcd(n+2, m+2); 67 init(); 68 Matrix_mod(pos); 69 } 70 return 0; 71 }
