A:
直接打表所有可以到达的点就可以了
1 #include <math.h>
2 #include <stdio.h>
3 #include <stdlib.h>
4 #include <iostream>
5 #include <algorithm>
6 #include <string>
7 #include <string.h>
8 #include <vector>
9 #include <map>
10 #include <stack>
11 #include <set>
12 #include <queue>
13
14 #define LL long long
15 #define INF 0x3f3f3f3f
16 #define ls nod<<1
17 #define rs (nod<<1)+1
18
19 const int maxn = 5e5 + 10;
20 const int MOD = 433494437;
21
22 template<class T>inline void read(T &res)
23 {
24 char c;T flag=1;
25 while((c=getchar())<'0'||c>'9')if(c=='-')flag=-1;res=c-'0';
26 while((c=getchar())>='0'&&c<='9')res=res*10+c-'0';res*=flag;
27 }
28
29 std::set<int> s1,s2;
30 int x,y,z;
31
32 int main() {
33 read(x);read(y);read(z);
34 s1.insert(0);
35 for (int i = 1;i <= 12;i++) {
36 for (auto it : s1) {
37 s2.insert(it + x);
38 s2.insert(it - x);
39 s2.insert(it + y);
40 s2.insert(it - y);
41 s2.insert(it + z);
42 s2.insert(it - z);
43 }
44 s1 = s2;
45 s2.clear();
46 }
47 int q;
48 int now = 0;
49 read(q);
50 while (q--) {
51 int temp;
52 read(temp);
53 if (s1.count(temp-now)) {
54 printf("YES\n");
55 now = abs(temp-now);
56 }
57 else {
58 printf("NO\n");
59 now = 0;
60 }
61 }
62 return 0;
63 }
B：
分别统计行和列的磨损度然后取最大
1 #include <math.h>
2 #include <stdio.h>
3 #include <stdlib.h>
4 #include <iostream>
5 #include <algorithm>
6 #include <string>
7 #include <string.h>
8 #include <vector>
9 #include <map>
10 #include <stack>
11 #include <set>
12 #include <queue>
13
14 #define LL long long
15 #define INF 0x3f3f3f3f
16 #define ls nod<<1
17 #define rs (nod<<1)+1
18
19 const int maxn = 1e5 + 10;
20 const int MOD = 433494437;
21
22 LL row[maxn],col[maxn];
23
24 int main() {
25 int n,m;
26 while (~scanf("%d%d",&n,&m)) {
27 memset(row,0, sizeof(row));
28 memset(col,0, sizeof(col));
29 for (int i = 1;i <= n;i++) {
30 for (int j = 1;j <= m;j++) {
31 int key;
32 scanf("%d",&key);
33 row[i] += key;
34 col[j] += key;
35 }
36 }
37 LL ans = 0;
38 for (int i = 1;i <= n;i++)
39 ans = std::max(row[i],ans);
40 for (int i = 1;i <= m;i++)
41 ans = std::max(col[i],ans);
42 printf("%lld\n",ans);
43 }
44 return 0;
45 }
C：
审题注意一下只需要跑一次的BFS就可以了，然后就是很简单的方向BFS。最后进行一个结构体排序就好了
1 #include <math.h>
2 #include <stdio.h>
3 #include <stdlib.h>
4 #include <iostream>
5 #include <algorithm>
6 #include <string>
7 #include <string.h>
8 #include <vector>
9 #include <map>
10 #include <stack>
11 #include <set>
12 #include <queue>
13
14 #define LL long long
15 #define INF 0x3f3f3f3f
16 #define ls nod<<1
17 #define rs (nod<<1)+1
18
19 const int maxn = 5e5 + 10;
20 const int MOD = 433494437;
21
22 template<class T>inline void read(T &res)
23 {
24 char c;T flag=1;
25 while((c=getchar())<'0'||c>'9')if(c=='-')flag=-1;res=c-'0';
26 while((c=getchar())>='0'&&c<='9')res=res*10+c-'0';res*=flag;
27 }
28
29 int n,m,cnt;
30 int dir[4][2] = {{-1,0},{1,0},{0,-1},{0,1}};
31 int vis[105][105];
32 char map[105][105];
33
34 struct Node {
35 int x;
36 int y;
37 }node[maxn];
38
39 bool cmp(Node a,Node b) {
40 if (a.x != b.x)
41 return a.x < b.x;
42 else
43 return a.y < b.y;
44 }
45
46 void dfs(int x,int y) {
47 for (int i = 0;i < 4;i++) {
48 int x1 = x + dir[i][0];
49 int y1 = y + dir[i][1];
50 if (!vis[x1][y1] && map[x1][y1] == '*' && x1 >= 1 && x1 <= n && y1 >= 1 && y1 <= m) {
51 vis[x1][y1] = 1;
52 continue;
53 }
54 if (vis[x1][y1] == 1 && map[x1][y1] == '*') {
55 vis[x1][y1] = 2;
56 node[cnt].x = x1;
57 node[cnt++].y = y1;
58 }
59 }
60 }
61
62
63 int main() {
64 while (~scanf("%d%d",&n,&m)) {
65 memset(vis,0, sizeof(vis));
66 cnt = 0;
67 for (int i = 1;i <= n;i++)
68 scanf("%s",map[i]+1);
69 for (int i = 1;i <= n;i++) {
70 for (int j = 1;j <= m;j++) {
71 if (map[i][j] == '#')
72 dfs(i,j);
73 }
74 }
75 if (cnt == 0) {
76 printf("-1\n");
77 continue;
78 }
79 std::sort(node,node+cnt,cmp);
80 for (int i = 0;i < cnt;i++)
81 printf("%d %d\n",node[i].x,node[i].y);
82 }
83 return 0;
84 }
D：
前几天刚做了一个和这个类似的题目。
对于求最小/大的符合条件的值的题目，我们都是采取二分答案的做法然后判断是否符合条件就可以了
这题就是典型的二分
1 #include <math.h>
2 #include <stdio.h>
3 #include <stdlib.h>
4 #include <iostream>
5 #include <algorithm>
6 #include <string>
7 #include <string.h>
8 #include <vector>
9 #include <map>
10 #include <stack>
11 #include <set>
12 #include <queue>
13
14 #define LL long long
15 #define INF 0x3f3f 3f3f
16 #define ls nod<<1
17 #define rs (nod<<1)+1
18
19 const int maxn = 2e6 + 10;
20 const int MOD = 433494437;
21
22 int n,m;
23 int a[maxn];
24
25 bool check(int k) {
26 int time = 0;
27 for (int i = 1;i <= n;i++) {
28 if (a[i] % k == 0)
29 time += (a[i] / k);
30 else
31 time += (a[i] /k + 1);
32 if (time > m)
33 return false;
34 }
35 return time<=m;
36 }
37
38
39 int main() {
40 while (~scanf("%d%d",&n,&m)) {
41 int Max = 0;
42 for (int i = 1;i <= n;i++) {
43 scanf("%d", &a[i]);
44 Max = std::max(a[i],Max);
45 }
46 int l = 1,r = Max;
47 int ans = Max;
48 while (l <= r) {
49 int mid = (l + r) >> 1;
50 if (check(mid)) {
51 ans = mid;
52 r = mid - 1;
53 }
54 else
55 l = mid + 1;
56 }
57 printf("%d\n",ans);
58 }
59 return 0;
60 }
E：
sort就完事了
F：
这道题的规律其实很好找。 最后的答案就是 2^x * (每层容积) 求和
然后注意一下循环没有必要循环到真正的 1e15 （n的范围） ，因为阶层的增长的速度很快，我们打表发现 178！% mod 此时已经是0了，所以超过178的它的值就是0
1 #include <math.h>
2 #include <stdio.h>
3 #include <stdlib.h>
4 #include <iostream>
5 #include <algorithm>
6 #include <string>
7 #include <string.h>
8 #include <vector>
9 #include <map>
10 #include <stack>
11 #include <set>
12 #include <queue>
13
14 #define LL long long
15 #define INF 0x3f3f3f3f
16 #define ls nod<<1
17 #define rs (nod<<1)+1
18
19 const int maxn = 5e5 + 10;
20 const LL MOD = 3777105087;
21
22 template<class T>inline void read(T &res)
23 {
24 char c;T flag=1;
25 while((c=getchar())<'0'||c>'9')if(c=='-')flag=-1;res=c-'0';
26 while((c=getchar())>='0'&&c<='9')res=res*10+c-'0';res*=flag;
27 }
28
29 LL a[maxn],d[maxn];
30
31 void pre() {
32 a[0] = 1;
33 for (int i = 1;i < 11;i++) {
34 a[i] = 2 * a[i-1];
35 }
36 }
37
38 void jie() {
39 d[1] = 1;
40 for (int i = 2;i <= 178;i++) {
41 d[i] = (i * d[i-1] ) % MOD;
42 }
43 }
44
45 int main() {
46 jie();
47 pre();
48 LL n;
49 while (~scanf("%lld",&n)) {
50 int x,y;
51 scanf("%d%d",&x,&y);
52 n = std::min(178ll,n);
53 LL sum = 0;
54 for (int i = 2;i <= n;i++) {
55 sum += d[i];
56 sum %= MOD;
57 }
58 if (y) {
59 sum = (sum * a[x]) % MOD;
60 }
61 sum %= MOD;
62 printf("%lld\n",sum+1);
63 }
64 return 0;
65 }
G：
这场比赛G题是一个好题。
我们可以先进行下分类：1、长度超过m的肯定是符合条件的 2、长度等于m的部分符合条件
对于第一种情况：
我们可以用组合数的方式进行计算就好了（注意前导‘0’的情况）
对于第二种情况：
我们很容易想到采用dp的方式去优化
这里我们让 dp[i][j]代表 第一个字符串到 i 位置、第二个字符串取前 j 个的时候值相等的个数
转移方程：
因为我们统计的只是相等的情况，那么其实对于 ch1[i] > ch2[j] 或者 ch1[i] < ch2[j] 这两种情况我们直接把前面的转移过来就可以了 dp[i][j] = dp[i-1][j]
对于 ch1[i] = ch2[j] ，dp[i][j] = dp[i-1][j] + dp[i-1][j-1]
（不管此时的位置） （加上此时位置）
当 ch1[i] > ch2[j] 的时候我们可以利用组合数去计算 （dp[i-1][j-1] （前面相等的部分）在加上这大于的两个(ch1[i]、ch2[j] ) 后面的话就是自由组合了 ）
1 #include <math.h>
2 #include <stdio.h>
3 #include <stdlib.h>
4 #include <iostream>
5 #include <algorithm>
6 #include <string>
7 #include <string.h>
8 #include <vector>
9 #include <map>
10 #include <stack>
11 #include <set>
12 #include <queue>
13
14 #define LL long long
15 #define INF 0x3f3f3f3f
16 #define ls nod<<1
17 #define rs (nod<<1)+1
18
19 const int maxn = 5e5 + 10;
20 const LL MOD = 998244353;
21
22 template<class T>inline void read(T &res)
23 {
24 char c;T flag=1;
25 while((c=getchar())<'0'||c>'9')if(c=='-')flag=-1;res=c-'0';
26 while((c=getchar())>='0'&&c<='9')res=res*10+c-'0';res*=flag;
27 }
28
29 char p[maxn],s[maxn];
30
31
32 LL C[3010][3010];//G++ long long
33 LL dp[3010][3010];
34 void init() {
35 C[0][0] = 1;
36 for(int i = 1; i < 3010; i++) {
37 C[i][0] = 1;
38 for(int j = 1; j <= i; j++) {
39 C[i][j] = (C[i - 1][j] + C[i - 1][j - 1]) % MOD;
40 }
41 }
42 }
43 int main() {
44 int t;
45 scanf("%d",&t);
46 init();
47 while (t--) {
48 int n,m;
49 LL ans = 0;
50 scanf("%d%d",&n,&m);
51 scanf("%s",s+1);
52 scanf("%s",p+1);
53 for (int i = 1;i <= n && n - i >= m;i++) {
54 if (s[i] != '0') {
55 for (int k = m;k <= n-i;k++) {
56 ans = (ans + C[n-i][k] ) % MOD;
57 }
58 }
59 }
60 for (int i = 0;i <= n+1;i++)
61 dp[i][0] = 1;
62 for (int j = 1;j <= m;j++) {
63 for (int i = j;i <= n;i++) {
64 dp[i][j] = dp[i-1][j];
65 if (s[i] == p[j])
66 dp[i][j] = (dp[i-1][j] + dp[i-1][j-1]) % MOD;
67 else if (s[i] > p[j])
68 ans = (ans + (dp[i-1][j-1] * C[n-i][m-j]) % MOD) % MOD;
69 }
70 }
71 printf("%lld\n",ans);
72 }
73 return 0;
74 }
H：
根据题意进行模拟就好了
1 #include <math.h>
2 #include <stdio.h>
3 #include <stdlib.h>
4 #include <iostream>
5 #include <algorithm>
6 #include <string>
7 #include <string.h>
8 #include <vector>
9 #include <map>
10 #include <stack>
11 #include <set>
12 #include <queue>
13
14 #define LL long long
15 #define INF 0x3f3f 3f3f
16 #define ls nod<<1
17 #define rs (nod<<1)+1
18
19 const int maxn = 1e5 + 10;
20 const int MOD = 433494437;
21
22 int f[11];
23 int ch[11][520][520] = {0};
24
25 int main() {
26 int t;
27 f[1] = 1;
28 for (int i = 2;i <= 10;i++)
29 f[i] = 2 * f[i-1];
30 ch[1][1][1] = 1;
31 for (int i = 2;i <= 10;i++) {
32 for (int j = 1;j <= f[i-1];j++) {
33 for (int k = 1,l = 1;k <= f[i];k++,l++) {
34 if (l > f[i-1])
35 l = 1;
36 ch[i][j][k] = ch[i-1][j][l];
37 }
38 }
39 for (int j = f[i-1]+1,row = 1;j <= f[i];j++,row++) {
40 for (int k = 1;k <= f[i-1];k++) {
41 ch[i][j][k] = !ch[i-1][row][k];
42 }
43 for (int k = f[i-1]+1,l = 1;k <= f[i];k++,l++)
44 ch[i][j][k] = ch[i-1][row][l];
45 }
46 }
47 scanf("%d",&t);
48 while (t--) {
49 int n;
50 scanf("%d",&n);
51 for (int i = 1;i <= f[n];i++) {
52 for (int j = 1;j <= f[n];j++) {
53 printf("%d ",ch[n][i][j]);
54 }
55 printf("\n");
56 }
57 }
58 }
I：
利用二进制进行统计
1 #include <math.h>
2 #include <stdio.h>
3 #include <stdlib.h>
4 #include <iostream>
5 #include <algorithm>
6 #include <string>
7 #include <string.h>
8 #include <vector>
9 #include <map>
10 #include <stack>
11 #include <set>
12 #include <queue>
13
14 #define LL long long
15 #define INF 0x3f3f3f3f
16 #define ls nod<<1
17 #define rs (nod<<1)+1
18
19 const int maxn = 1e5 + 10;
20 const int MOD = 433494437;
21
22 int Func(LL val) {
23 LL m = 1;
24 int cnt = 0;
25 for (int i = 1;i < 51;i++) {
26 if ((m & val) == m) {
27 cnt++;
28 }
29 m = m << 1;
30 }
31 return cnt;
32 }
33
34 int main() {
35 LL k;
36 while (~scanf("%lld",&k)) {
37 printf("%d\n",Func(k));
38 }
39 return 0;
40 }
J：
题目的意思其实就是让你找最长的连续的个数
1 5 6 2 4 3
我们答案应该是 3
因为我们相当于找到最长连续的，这里是 1 2 3
转移方程 ：dp[a[i]] = dp[a[i]-1] + 1
1 #include <math.h>
2 #include <stdio.h>
3 #include <stdlib.h>
4 #include <iostream>
5 #include <algorithm>
6 #include <string>
7 #include <string.h>
8 #include <vector>
9 #include <map>
10 #include <stack>
11 #include <set>
12 #include <queue>
13
14 #define LL long long
15 #define INF 0x3f3f3f3f
16 #define ls nod<<1
17 #define rs (nod<<1)+1
18
19 const int maxn = 5e5 + 10;
20 const int MOD = 433494437;
21
22 template<class T>inline void read(T &res)
23 {
24 char c;T flag=1;
25 while((c=getchar())<'0'||c>'9')if(c=='-')flag=-1;res=c-'0';
26 while((c=getchar())>='0'&&c<='9')res=res*10+c-'0';res*=flag;
27 }
28 int a[maxn],dp[maxn];
29
30 int main() {
31 int t;
32 read(t);
33 while (t--) {
34 int n;
35 read(n);
36 int Max = 0;
37 for (int i = 1;i <= n;i++) {
38 read(a[i]);
39 dp[i] = 0;
40 }
41 for (int i = 1;i <= n;i++) {
42 dp[a[i]] = dp[a[i]-1] + 1;
43 Max = std::max(Max,dp[a[i]]);
44 }
45 printf("%d\n",n-Max);
46 }
47 return 0;
48 }
K：
冰查鸡
1 #include <math.h>
2 #include <stdio.h>
3 #include <stdlib.h>
4 #include <iostream>
5 #include <algorithm>
6 #include <string>
7 #include <string.h>
8 #include <vector>
9 #include <map>
10 #include <stack>
11 #include <set>
12 #include <queue>
13
14 #define LL long long
15 #define INF 0x3f3f3f3f
16 #define ls nod<<1
17 #define rs (nod<<1)+1
18
19 const int maxn = 5e5 + 10;
20 const LL MOD = 3777105087;
21
22 template<class T>inline void read(T &res)
23 {
24 char c;T flag=1;
25 while((c=getchar())<'0'||c>'9')if(c=='-')flag=-1;res=c-'0';
26 while((c=getchar())>='0'&&c<='9')res=res*10+c-'0';res*=flag;
27 }
28
29 int pre[1010],vis[1005];
30 int tong[1005][15];
31 char ch[1010];
32
33 int find(int x) {
34 if (pre[x] == x)
35 return x;
36 return pre[x] = find(pre[x]);
37 }
38
39 void merge(int x,int y) {
40 int rootx = find(x),rooty = find(y);
41 pre[rooty] = rootx;
42 }
43
44 int main() {
45 int t;
46 scanf("%d",&t);
47 while (t--) {
48 memset(tong, 0, sizeof(tong));
49 memset(vis,0, sizeof(vis));
50 memset(pre,0, sizeof(pre));
51 int n;
52 scanf("%d", &n);
53 for (int i = 1;i <= n;i++)
54 pre[i] = i;
55 for (int i = 1; i <= n; i++) {
56 memset(ch, '\0', sizeof(ch));
57 scanf("%s", ch);
58 for (int j = 0; j < strlen(ch); j++)
59 tong[i][ch[j] - '0'] = 1;
60 }
61 for (int i = 1; i <= n; i++)
62 for (int j = i + 1; j <= n; j++)
63 for (int l1 = 0; l1 <= 9; l1++)
64 if (tong[i][l1] == tong[j][l1] && tong[i][l1])
65 merge(i, j);
66 int cnt = 0;
67 for (int i = 1;i <= n;i++)
68 vis[find(i)] = 1;
69 for (int i = 1;i <= n;i++)
70 if (vis[i])
71 cnt++;
72 printf("%d\n",cnt);
73 }
74 return 0;
75 }
L：
大模拟
不符合规则的几种情况优先判断。
我们先放一个，然后进行判断是否胜利。然后再放对手的一个，判断对手是否胜利。最后自己再放一个，判断自己是否胜利。
1 #include <math.h>
2 #include <stdio.h>
3 #include <stdlib.h>
4 #include <iostream>
5 #include <algorithm>
6 #include <string>
7 #include <string.h>
8 #include <vector>
9 #include <map>
10 #include <stack>
11 #include <set>
12 #include <queue>
13
14 #define LL long long
15 #define INF 0x3f3f3f3f
16 #define ls nod<<1
17 #define rs (nod<<1)+1
18
19 const int maxn = 5e5 + 10;
20 const LL MOD = 3777105087;
21
22 template<class T>inline void read(T &res)
23 {
24 char c;T flag=1;
25 while((c=getchar())<'0'||c>'9')if(c=='-')flag=-1;res=c-'0';
26 while((c=getchar())>='0'&&c<='9')res=res*10+c-'0';res*=flag;
27 }
28
29 char map[5][5];
30
31 bool judge(char k) {
32 for (int i = 1;i <= 3;i++)
33 if (map[i][1] == k && map[i][2] == k && map[i][3] == k)
34 return true;
35 for (int i = 1;i <= 3;i++)
36 if (map[1][i] == k && map[2][i] == k && map[3][i] == k)
37 return true;
38 if (map[1][1] == k && map[2][2] == k && map[3][3] == k)
39 return true;
40 if (map[3][1] == k && map[2][2] == k && map[1][3] == k)
41 return true;
42 return false;
43 }
44
45 bool win_1(char k) {
46 for (int i = 1;i <= 3;i++) {
47 for (int j = 1;j <= 3;j++) {
48 if (map[i][j] == '.') {
49 map[i][j] = k;
50 if (judge(k)) {
51 map[i][j] = '.';
52 return true;
53 }
54 map[i][j] = '.';
55 }
56 }
57 }
58 return false;
59 }
60
61 bool win_2(char k) {
62 if (judge(k))
63 return true;
64 if (win_1(k == 'o' ? 'x' : 'o'))
65 return false;
66 for (int i = 1;i <= 3;i++) {
67 for (int j = 1;j <= 3;j++) {
68 if (map[i][j] == '.') {
69 if (k == 'o')
70 map[i][j] = 'x';
71 else
72 map[i][j] = 'o';
73 if (win_1(k))
74 map[i][j] = '.';
75 else {
76 map[i][j] = '.';
77 return false;
78 }
79 }
80 }
81 }
82 return true;
83
84 }
85
86 bool solve(char k) {
87 for (int i = 1;i <= 3;i++) {
88 for (int j = 1;j <= 3;j++) {
89 if (map[i][j] == '.') {
90 map[i][j] = k;
91 if (win_2(k))
92 return true;
93 map[i][j] = '.';
94 }
95 }
96 }
97 return false;
98 }
99
100 int main() {
101 int t;
102 std::cin >> t;
103 while (t--) {
104 int cnto = 0,cntx = 0;
105 for (int i = 1;i <= 3;i++) {
106 for (int j = 1;j <= 3;j++) {
107 std::cin >> map[i][j];
108 if (map[i][j] == 'o')
109 cnto++;
110 if (map[i][j] == 'x')
111 cntx++;
112 }
113 }
114 char temp;
115 std::cin >> temp;
116 if (temp == 'o') {
117 if (cnto > cntx) {
118 printf("wrong!\n");
119 continue;
120 }
121 }
122 if (temp == 'x') {
123 if (cntx > cnto) {
124 printf("wrong!\n");
125 continue;
126 }
127 }
128 if (abs(cnto - cntx) >= 2 || judge('o') || judge('x')) {
129 printf("wrong!\n");
130 continue;
131 }
132 if (win_1(temp)) {
133 printf("LeeLdler win!\n");
134 continue;
135 }
136 if (solve(temp)) {
137 printf("LeeLdler win!\n");
138 continue;
139 }
140 else
141 printf("Cannot win!\n");
142 }
143 return 0;
144 }
M：
从容积为0开始倒着往前推。
能直接到达0的肯定是先手必胜
1 #include <math.h>
2 #include <stdio.h>
3 #include <stdlib.h>
4 #include <iostream>
5 #include <algorithm>
6 #include <string>
7 #include <string.h>
8 #include <vector>
9 #include <map>
10 #include <stack>
11 #include <set>
12 #include <queue>
13
14 #define LL long long
15 #define INF 0x3f3f3f3f
16 #define ls nod<<1
17 #define rs (nod<<1)+1
18
19 const int maxn = 5e5 + 10;
20 const LL MOD = 3777105087;
21
22 template<class T>inline void read(T &res)
23 {
24 char c;T flag=1;
25 while((c=getchar())<'0'||c>'9')if(c=='-')flag=-1;res=c-'0';
26 while((c=getchar())>='0'&&c<='9')res=res*10+c-'0';res*=flag;
27 }
28
29 int a[maxn],d[maxn];
30
31 int main() {
32 int n,m;
33 read(n);read(m);
34 for (int i = 1;i <= m;i++)
35 read(a[i]);
36 for (int i = 0;i <= n;i++) {
37 for (int j = 1;j <= m;j++) {
38 if (i + a[j] > n)
39 continue;
40 if (d[i] == 0)
41 d[i+a[j]] = 1;
42 }
43 }
44 if (d[n])
45 printf("Y\n");
46 else
47 printf("N\n");
48 return 0;
49 }
来源：https://www.cnblogs.com/-Ackerman/p/12169175.html