【HDOJ】4541 Ten Googol
打表的大水题。
1 /* 4541 */ 2 #include <cstdio> 3 #include <cstdlib> 4 #include <cstring> 5 6 int a3[] = {1, 2, 6, 10}; 7 int a4[] = {0, 4, 5, 9}; 8 int a5[] = {3, 7, 8, 40, 50, 60}; 9 int a6[] = {11, 12, 20, 30, 80, 90}; 10 int a7[] = {15, 16, 70}; 11 int a8[] = {13, 14, 18, 19, 41, 42, 46, 51, 52, 56, 61, 62, 66}; 12 int a9[] = {17, 21, 22, 26, 31, 32, 36, 44, 45, 49, 54, 55, 59, 64, 65, 69, 81, 82, 86, 91, 92, 96}; 13 14 15 int main() { 16 int t, m, n; 17 int n3, n4, n5, n6, n7, n8, n9; 18 int i; 19 20 #ifndef ONLINE_JUDGE 21 freopen("data.in", "r", stdin); 22 freopen("data.out", "w", stdout); 23 #endif 24 25 n3 = sizeof(a3)/sizeof(int); 26 n4 = sizeof(a4)/sizeof(int); 27 n5 = sizeof(a5)/sizeof(int); 28 n6 = sizeof(a6)/sizeof(int); 29 n7 = sizeof(a7)/sizeof(int); 30 n8 = sizeof(a8)/sizeof(int); 31 n9 = sizeof(a9)/sizeof(int); 32 33 scanf("%d", &t); 34 for (i=1; i<=t; ++i) { 35 scanf("%d %d", &n, &m); 36 printf("Case #%d: ", i); 37 --m; 38 if (n == 3) { 39 if (m >= n3) 40 printf("-1\n"); 41 else 42 printf("%d\n", a3[m]); 43 } 44 if (n == 4) { 45 if (m >= n4) 46 printf("-1\n"); 47 else 48 printf("%d\n", a4[m]); 49 } 50 if (n == 5) { 51 if (m >= n5) 52 printf("-1\n"); 53 else 54 printf("%d\n", a5[m]); 55 } 56 if (n == 6) { 57 if (m >= n6) 58 printf("-1\n"); 59 else 60 printf("%d\n", a6[m]); 61 } 62 if (n == 7) { 63 if (m >= n7) 64 printf("-1\n"); 65 else 66 printf("%d\n", a7[m]); 67 } 68 if (n == 8) { 69 if (m >= n8) 70 printf("-1\n"); 71 else 72 printf("%d\n", a8[m]); 73 } 74 if (n == 9) { 75 if (m < n9) 76 printf("%d\n", a9[m]); 77 else if (m == n9) 78 printf("10000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000\n"); 79 else if (m == n9+1) 80 printf("20000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000\n"); 81 else if (m == n9+2) 82 printf("60000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000\n"); 83 else if (m == n9+3) 84 printf("100000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000\n"); 85 else 86 printf("-1\n"); 87 } 88 } 89 90 return 0; 91 }
