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Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2175 Accepted Submission(s): 1110
1 #include <iostream> 2 #include <cstring> 3 #include <string> 4 #include <cstdio> 5 #include <cmath> 6 7 using namespace std; 8 9 #define MOD 2 10 #define MAXN 101 11 12 typedef struct MAT 13 { 14 int d[MAXN][MAXN]; 15 int r, c; 16 MAT() 17 { 18 r = c = 0; 19 memset(d, 0, sizeof(d)); 20 } 21 }MAT; 22 23 MAT mul(MAT m1, MAT m2, int mod) 24 { 25 MAT ans = MAT(); 26 ans.r = m1.r; 27 ans.c = m2.c; 28 for(int i = 0; i < m1.r; i++) 29 { 30 for(int j = 0; j < m2.r; j++) 31 { 32 if(m1.d[i][j]) 33 { 34 for(int k = 0; k < m2.c; k++) 35 { 36 ans.d[i][k] = (ans.d[i][k] + m1.d[i][j] * m2.d[j][k]) % mod; 37 } 38 } 39 } 40 } 41 return ans; 42 } 43 44 MAT quickmul(MAT m, int n, int mod) 45 { 46 MAT ans = MAT(); 47 for(int i = 0; i < m.r; i++) 48 { 49 ans.d[i][i] = 1; 50 } 51 ans.r = m.r; 52 ans.c = m.c; 53 while(n) 54 { 55 if(n & 1) 56 { 57 ans = mul(m, ans, mod); 58 } 59 m = mul(m, m, mod); 60 n >>= 1; 61 } 62 return ans; 63 } 64 65 int main() { 66 int t; 67 while(scanf("%d", &t) != EOF && t) { 68 char T[105]; 69 scanf("%s", T); 70 int n = strlen(T); 71 MAT A, B, ans; 72 A.r = 1, A.c = n; 73 B.r = n, B.c = n; 74 for(int i = 0; i < n; i++) { 75 A.d[0][i] = T[i] - ‘0‘; 76 B.d[i][i] = 1; 77 B.d[i][(i+1)%n] = 1; 78 } 79 ans = quickmul(B, t, MOD); 80 ans = mul(A, ans, MOD); 81 for(int i = 0; i < n; i++) { 82 printf("%d", ans.d[0][i]); 83 } 84 printf("\n"); 85 } 86 return 0; 87 }
[HDOJ2276]Kiki & Little Kiki 2
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原文地址:http://www.cnblogs.com/vincentX/p/4579022.html