第三次月赛题解
1000
每天只要复习收益最大的那门课即可
1 #include <stdio.h> 2 #include <string.h> 3 #include <stdlib.h> 4 #include <algorithm> 5 #include <iostream> 6 #include <queue> 7 #include <stack> 8 #include <vector> 9 #include <map> 10 #include <set> 11 #include <string> 12 #include <math.h> 13 using namespace std; 14 #pragma warning(disable:4996) 15 typedef long long LL; 16 const int INF = 1<<30; 17 /* 18 */ 19 const int N = 100 + 10; 20 int dp[N][N]; 21 int a[N][N]; 22 char in[] = "D:\\3\\1004\\0.in"; 23 char out[] = "D:\\3\\1004\\0.out"; 24 char str[] = "0123456789"; 25 int main() 26 { 27 //for (int z = 0; z <= 9; ++z) 28 { 29 //in[10] = str[z]; 30 //out[10] = str[z]; 31 //freopen(in, "r", stdin); 32 //freopen(out, "w", stdout); 33 int n, m; 34 while (scanf("%d%d", &n, &m) != EOF) 35 { 36 for (int i = 1; i <= n; ++i) 37 { 38 for (int j = 1; j <= m; ++j) 39 scanf("%1d", &a[i][j]); 40 } 41 42 int ans = 0; 43 for (int i = 1; i <= n; ++i) 44 { 45 sort(a[i] + 1, a[i] + n + 1); 46 ans += a[i][n]; 47 } 48 printf("%d\n", ans); 49 } 50 } 51 return 0; 52 }
1001
个位数的位数是1，十位的是2， 所以只要算出个位数的个数*1，加上十位数的个数*2，加....
1 #include <stdio.h> 2 #include <string.h> 3 #include <stdlib.h> 4 #include <algorithm> 5 #include <iostream> 6 #include <queue> 7 #include <stack> 8 #include <vector> 9 #include <map> 10 #include <set> 11 #include <string> 12 #include <math.h> 13 using namespace std; 14 #pragma warning(disable:4996) 15 typedef __int64 LL; 16 const int INF = 1 << 30; 17 /* 18 19 */ 20 char in[] = "D:\\3\\1000\\0.in"; 21 char out[] = "D:\\3\\1000\\0.out"; 22 char str[] = "0123456789"; 23 int main() 24 { 25 //for (int z = 0; z <= 9; ++z) 26 { 27 //in[10] = str[z]; 28 //out[10] = str[z]; 29 //freopen(in, "r", stdin); 30 //freopen(out, "w", stdout); 31 LL n; 32 while (scanf("%I64d", &n) != EOF) 33 { 34 char str[11]; 35 LL m = 1; 36 sprintf(str, "%I64d", n); 37 LL len = strlen(str); 38 LL ans = 0; 39 LL t = 9; 40 LL i = 0; 41 for (i = 0; i < len - 1; ++i) 42 { 43 ans = ans + t * (i + 1); 44 t *= 10;//位数为i+1的数字有多少个 45 m = m * 10; 46 } 47 ans += (i + 1) * (n - m + 1); 48 printf("%I64d\n", ans); 49 //puts("8888888899"); 50 } 51 } 52 53 return 0; 54 }
1002
ABC三根柱子
设T(n)是将n个盘子借助一根柱子移到另一个柱子上去所要移动的最小次数， 那么要先将n-1个盘子移到B柱子上去，需要T(n-1)次，然后将第n个盘子移到C柱子上去，需要一次，然后将n-1个盘子移动C柱子上去
需要T(n-1)次， 所以T(n) = 2*T(n-1) + 1
设T(n)+1 = 2( T(n-1)+1 ) = 2 * 2( T(n-2)+1) = ... = 2^n 所以T(n) = 2^n - 1
需要注意的是2^64 会爆
1 #include <stdio.h> 2 #include <string.h> 3 #include <stdlib.h> 4 #include <algorithm> 5 #include <iostream> 6 #include <queue> 7 #include <stack> 8 #include <vector> 9 #include <map> 10 #include <set> 11 #include <string> 12 #include <math.h> 13 using namespace std; 14 #pragma warning(disable:4996) 15 typedef unsigned __int64 LL; 16 const int INF = 1<<30; 17 /* 18 19 */ 20 char str[] = "0123456789"; 21 char addressIn[] = "D:\\3\\1001\\1.in"; 22 char addressOut[] = "D:\\3\\1001\\1.out"; 23 int main() 24 { 25 int n; 26 LL ans; 27 //for (int i = 0; i < 9; ++i) 28 { 29 //addressIn[10] = str[i]; 30 //addressOut[10] = str[i]; 31 32 //freopen(addressIn, "r", stdin); 33 //freopen(addressOut, "w", stdout); 34 while (scanf("%d", &n) != EOF) 35 { 36 ans = 0; 37 if (n == 0) 38 { 39 puts("0"); 40 continue; 41 } 42 else if (n == 64) 43 ans = 0; 44 else 45 { 46 ans = 1; 47 ans <<= n; 48 } 49 ans -= 1; 50 printf("%I64u\n", ans); 51 } 52 } 53 return 0; 54 }
1003
根据异或的性质， a^a = 0 a^a^a = a , 所以将n个数字进行异或， 如果结果是0，说明所有的数字都出现了偶数次， 如果结果不为0，那么就输出
需要注意的是，如果a为0，切出现了奇数次， 其异或结果也是0 ， 所以异或时将每个数字加1即可
1 #include <stdio.h> 2 #include <string.h> 3 #include <stdlib.h> 4 #include <algorithm> 5 #include <iostream> 6 #include <queue> 7 #include <stack> 8 #include <vector> 9 #include <map> 10 #include <set> 11 #include <string> 12 #include <math.h> 13 #include<time.h> 14 using namespace std; 15 #pragma warning(disable:4996) 16 typedef long long LL; 17 const int INF = 1<<30; 18 /* 19 20 */ 21 __int64 input() 22 { 23 char ch = getchar(); 24 while (ch<'0' || ch>'9') 25 ch = getchar(); 26 __int64 x = 0; 27 while (ch >= '0' && ch <= '9') 28 { 29 x = x * 10 + ch - '0'; 30 ch = getchar(); 31 } 32 return x; 33 } 34 const LL MAX = 1152921504606846976LL; 35 LL a[5000000+10]; 36 char in[] = "D:\\3\\1002\\0.in"; 37 char out[] = "D:\\3\\1002\\0.out"; 38 char str[] = "0123456789"; 39 int main() 40 { 41 int n; 42 //for (int k = 0; k <= 9; ++k) 43 { 44 //in[10] = str[k]; 45 //freopen(in, "r", stdin); 46 //freopen("D:\\3\\1002\\0.out", "w", stdout); 47 48 while (scanf("%d",&n)!=EOF) 49 { 50 LL ans = 0, x; 51 for (int i = 0; i < n; ++i) 52 { 53 //scanf("%I64d", &a[i]); 54 //input(a[i]); 55 a[i] = input(); 56 a[i]++; 57 ans ^= a[i]; 58 } 59 //sort(a, a + n); 60 printf("%I64d\n", ans==0?-1:ans-1); 61 } 62 } 63 return 0; 64 }
1004
这个问题难点是不知道要打扫哪几列， 其实只要我们枚举每一行，使得这一行全为1， 那么就知道了要打扫哪几列， 然后判断剩下的n-1行，看打扫了这几列之后是否全为1
1 #include <stdio.h> 2 #include <string.h> 3 #include <stdlib.h> 4 #include <algorithm> 5 #include <iostream> 6 #include <queue> 7 #include <stack> 8 #include <vector> 9 #include <map> 10 #include <set> 11 #include <string> 12 #include <math.h> 13 using namespace std; 14 #pragma warning(disable:4996) 15 typedef long long LL; 16 const int INF = 1 << 30; 17 /* 18 19 */ 20 const int N = 101; 21 char a[N][N]; 22 int ans; 23 int vis[N]; 24 char in[] = "D:\\3\\1005\\9.in"; 25 char out[] = "D:\\3\\1005\\9.out"; 26 char str[] = "0123456789"; 27 int main() 28 { 29 30 //for (int z = 0; z <= 9; ++z) 31 { 32 //in[10] = str[z]; 33 //out[10] = str[z]; 34 //freopen(in, "r", stdin); 35 //freopen(out, "w", stdout); 36 int n; 37 while (scanf("%d", &n) != EOF) 38 { 39 memset(vis, 0, sizeof(vis)); 40 for (int i = 0; i < n; ++i) 41 scanf("%s", a[i]); 42 43 ans = 0; 44 for (int i = 0; i < n; ++i) 45 { 46 bool flag = true; 47 for (int j = 0; j < n; ++j) 48 if (a[i][j] == '0') 49 { 50 flag = false; 51 break; 52 } 53 if (flag) 54 ans++; 55 } 56 int tmp; 57 for (int i = 0; i < n; ++i) 58 { 59 memset(vis, 0, sizeof(vis)); 60 for (int j = 0; j < n; ++j) 61 { 62 if (a[i][j] == '0') 63 vis[j] = true; 64 } 65 tmp = 1; 66 for (int j = 0; j < n; ++j) 67 { 68 if (i == j) continue; 69 bool f = true; 70 for (int k = 0; k < n; ++k) 71 if (a[j][k] == '0' && !vis[k]) 72 { 73 f = false; 74 break; 75 } 76 else if (a[j][k] == '1' && vis[k]) 77 { 78 f = false; 79 break; 80 } 81 if (f) 82 tmp++; 83 } 84 ans = ans > tmp ? ans : tmp; 85 } 86 printf("%d\n", ans); 87 } 88 } 89 return 0; 90 }
1005
