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
时间： 2024-11-15 00:45:50