第三次月赛题解-白红宇

第三次月赛题解

阅读量：5339 次

发布时间：2019-06-15

本文共 6358 字，大约阅读时间需要 21 分钟。

1000

每天只要复习收益最大的那门课即可

1 #include 
      
        2 #include 
       
         3 #include 
        
          4 #include 
         
           5 #include 
          
            6 #include 
           
             7 #include 
            
              8 #include 
             
               9 #include 
              10 #include 
               
                11 #include 
                
                 12 #include 
                 
                  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 }

View Code

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 
      
        2 #include 
       
         3 #include 
        
          4 #include 
         
           5 #include 
          
            6 #include 
           
             7 #include 
            
              8 #include 
             
               9 #include 
              10 #include 
               
                11 #include 
                
                 12 #include 
                 
                  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 }

View Code

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 
      
        2 #include 
       
         3 #include 
        
          4 #include 
         
           5 #include 
          
            6 #include 
           
             7 #include 
            
              8 #include 
             
               9 #include 
              10 #include 
               
                11 #include 
                
                 12 #include 
                 
                  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 else45 {46 ans = 1;47 ans <<= n;48 }49 ans -= 1;50 printf("%I64u\n", ans);51 }52 }53 return 0;54 }

View Code

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 
      
        2 #include 
       
         3 #include 
        
          4 #include 
         
           5 #include 
          
            6 #include 
           
             7 #include 
            
              8 #include 
             
               9 #include 
              10 #include 
               
                11 #include 
                
                 12 #include 
                 
                  13 #include
                  
                   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 }

View Code

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 
      
        2 #include 
       
         3 #include 
        
          4 #include 
         
           5 #include 
          
            6 #include 
           
             7 #include 
            
              8 #include 
             
               9 #include 
              10 #include 
               
                11 #include 
                
                 12 #include 
                 
                  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 }

View Code

1005

转载于:https://www.cnblogs.com/justPassBy/p/4604001.html

你可能感兴趣的文章