A. Payment Without Change
Description
给出a个价值为n的硬币和b个价值为1的硬币，问凑出来s元钱。
Solution
$/lfloor s/n /rfloor /times n + b \geq s$
我还憨憨写了个二分。结果只是因为爆int。
1 #include <algorithm>
2 #include <numeric>
3 #include <cctype>
4 #include <cmath>
5 #include <cstdio>
6 #include <cstdlib>
7 #include <cstring>
8 #include <iostream>
9 #include <map>
10 #include <queue>
11 #include <set>
12 #include <stack>
13 #if __cplusplus >= 201103L
14 #include <unordered_map>
15 #include <unordered_set>
16 #endif
17 #include <vector>
18 #define lson rt << 1, l, mid
19 #define rson rt << 1 | 1, mid + 1, r
20 #define LONG_LONG_MAX 9223372036854775807LL
21 #define pblank putchar(' ')
22 #define ll LL
23 #define fastIO ios::sync_with_stdio(false), cin.tie(0), cout.tie(0)
24 using namespace std;
25 typedef long long ll;
26 typedef long double ld;
27 typedef unsigned long long ull;
28 typedef pair<int, int> P;
29 int n, m, k;
30 const int maxn = 1e5 + 10;
31 template <class T>
32 inline T read()
33 {
34 int f = 1;
35 T ret = 0;
36 char ch = getchar();
37 while (!isdigit(ch))
38 {
39 if (ch == '-')
40 f = -1;
41 ch = getchar();
42 }
43 while (isdigit(ch))
44 {
45 ret = (ret << 1) + (ret << 3) + ch - '0';
46 ch = getchar();
47 }
48 ret *= f;
49 return ret;
50 }
51 template <class T>
52 inline void write(T n)
53 {
54 if (n < 0)
55 {
56 putchar('-');
57 n = -n;
58 }
59 if (n >= 10)
60 {
61 write(n / 10);
62 }
63 putchar(n % 10 + '0');
64 }
65 template <class T>
66 inline void writeln(const T &n)
67 {
68 write(n);
69 puts("");
70 }
71 template <typename T>
72 void _write(const T &t)
73 {
74 write(t);
75 }
76 template <typename T, typename... Args>
77 void _write(const T &t, Args... args)
78 {
79 write(t), pblank;
80 _write(args...);
81 }
82 template <typename T, typename... Args>
83 inline void write_line(const T &t, const Args &... data)
84 {
85 _write(t, data...);
86 }
87 int main(int argc, char const *argv[])
88 {
89 #ifndef ONLINE_JUDGE
90 freopen("in.txt","r", stdin);
91 // freopen("out.txt","w", stdout);
92 #endif
93 int t = read<int>();
94 while(t--){
95 ll a = read<int>(), b = read<int>(), n = read<int>(), s = read<int>();
96 ll l = 0, r = a;
97 ll res = -1;
98 while(l<=r){
99 ll mid = l + r>>1;
100 if (mid*n>s)
101 res = mid, r = mid - 1;
102 else
103 l = mid + 1;
104 }
105 if (res==-1){
106 if (a*n+b>=s)
107 puts("YES");
108 else
109 puts("NO");
110 }
111 else{
112 if ((res-1)*n+b>=s)
113 puts("YES");
114 else
115 puts("NO");
116 }
117 }
118 return 0;
119 }
B. Minimize the Permutation
Description
给出一个长为n的序列，一共可以进行n-1个操作，操作$i$可以将序列$a[j],a[j+1]$交换。
求交换后字典序最小的序列。
Solution
卡了很久这个，想想真的憨，后来发现是忘了更新pos值。
每次将最小的值移到尽可能的前面，记录操作了哪些步骤，判断满足大小关系或者当前步骤已经操作过则break
1 #include <algorithm>
2 #include <cctype>
3 #include <cmath>
4 #include <cstdio>
5 #include <cstdlib>
6 #include <cstring>
7 #include <iostream>
8 #include <map>
9 #include <numeric>
10 #include <queue>
11 #include <set>
12 #include <stack>
13 #if __cplusplus >= 201103L
14 #include <unordered_map>
15 #include <unordered_set>
16 #endif
17 #include <vector>
18 #define lson rt << 1, l, mid
19 #define rson rt << 1 | 1, mid + 1, r
20 #define LONG_LONG_MAX 9223372036854775807LL
21 #define pblank putchar(' ')
22 #define ll LL
23 #define fastIO ios::sync_with_stdio(false), cin.tie(0), cout.tie(0)
24 using namespace std;
25 typedef long long ll;
26 typedef long double ld;
27 typedef unsigned long long ull;
28 typedef pair<int, int> P;
29 int n, m, k;
30 const int maxn = 1e5 + 10;
31 template <class T>
32 inline T read()
33 {
34 int f = 1;
35 T ret = 0;
36 char ch = getchar();
37 while (!isdigit(ch))
38 {
39 if (ch == '-')
40 f = -1;
41 ch = getchar();
42 }
43 while (isdigit(ch))
44 {
45 ret = (ret << 1) + (ret << 3) + ch - '0';
46 ch = getchar();
47 }
48 ret *= f;
49 return ret;
50 }
51 template <class T>
52 inline void write(T n)
53 {
54 if (n < 0)
55 {
56 putchar('-');
57 n = -n;
58 }
59 if (n >= 10)
60 {
61 write(n / 10);
62 }
63 putchar(n % 10 + '0');
64 }
65 template <class T>
66 inline void writeln(const T &n)
67 {
68 write(n);
69 puts("");
70 }
71 template <typename T>
72 void _write(const T &t)
73 {
74 write(t);
75 }
76 template <typename T, typename... Args>
77 void _write(const T &t, Args... args)
78 {
79 write(t), pblank;
80 _write(args...);
81 }
82 template <typename T, typename... Args>
83 inline void write_line(const T &t, const Args &... data)
84 {
85 _write(t, data...);
86 }
87 int a[maxn], pos[maxn],vis[maxn];
88 int main(int argc, char const *argv[])
89 {
90 #ifndef ONLINE_JUDGE
91 freopen("in.txt", "r", stdin);
92 // freopen("out.txt","w", stdout);
93 #endif
94 int t = read<int>();
95 while (t--)
96 {
97 n = read<int>();
98 for (int i = 1; i <= n; i++)
99 {
100 a[i] = read<int>();
101 pos[a[i]] = i;
102 vis[i] = 0;
103 }
104 int left = n - 1;
105 pos[0] = 0;
106 for (int i = 1; i <n && left; i++)
107 {
108 for (int j = pos[i]-1; j >=1;j--){
109 if (a[j]<a[j+1]||vis[j])
110 break;
111 --left;
112 swap(a[j], a[j+1]);
113 swap(pos[a[j]], pos[a[j+1]]);
114 vis[j] = 1;
115 }
116 }
117 for (int i = 1; i <= n; i++)
118 write(a[i]), pblank;
119 puts("");
120 }
121 return 0;
122 }
C. Platforms Jumping
Description
给出长为n的一条河，以及m个不同长度木板。
小明每次最多可以跳p个单位长度。
问如何安排木板能使小明从0位置安全跳到n+1。
要求木板的交换顺序。
Solution
能否安全跳到最右边很好判定，不过如何安排木板就憨了。
先说一句dyznb。
首先考虑m块木板的总长度为sum，left为n-sum。
即left为小明需要跳跃的距离。
m块木板，加上0和n+1两个点。
一共有m+1个间隙，每个间隙平均分配$left/m$
如果有余数则将前面的部分加一个空隙。
注意判断边界。
1 #include <algorithm>
2 #include <cctype>
3 #include <cmath>
4 #include <cstdio>
5 #include <cstdlib>
6 #include <cstring>
7 #include <iostream>
8 #include <map>
9 #include <numeric>
10 #include <queue>
11 #include <set>
12 #include <stack>
13 #if __cplusplus >= 201103L
14 #include <unordered_map>
15 #include <unordered_set>
16 #endif
17 #include <vector>
18 #define lson rt << 1, l, mid
19 #define rson rt << 1 | 1, mid + 1, r
20 #define LONG_LONG_MAX 9223372036854775807LL
21 #define pblank putchar(' ')
22 #define ll LL
23 #define fastIO ios::sync_with_stdio(false), cin.tie(0), cout.tie(0)
24 using namespace std;
25 typedef long long ll;
26 typedef long double ld;
27 typedef unsigned long long ull;
28 typedef pair<int, int> P;
29 int n, m, k;
30 const int maxn = 1e5 + 10;
31 template <class T>
32 inline T read()
33 {
34 int f = 1;
35 T ret = 0;
36 char ch = getchar();
37 while (!isdigit(ch))
38 {
39 if (ch == '-')
40 f = -1;
41 ch = getchar();
42 }
43 while (isdigit(ch))
44 {
45 ret = (ret << 1) + (ret << 3) + ch - '0';
46 ch = getchar();
47 }
48 ret *= f;
49 return ret;
50 }
51 template <class T>
52 inline void write(T n)
53 {
54 if (n < 0)
55 {
56 putchar('-');
57 n = -n;
58 }
59 if (n >= 10)
60 {
61 write(n / 10);
62 }
63 putchar(n % 10 + '0');
64 }
65 template <class T>
66 inline void writeln(const T &n)
67 {
68 write(n);
69 puts("");
70 }
71 template <typename T>
72 void _write(const T &t)
73 {
74 write(t);
75 }
76 template <typename T, typename... Args>
77 void _write(const T &t, Args... args)
78 {
79 write(t), pblank;
80 _write(args...);
81 }
82 template <typename T, typename... Args>
83 inline void write_line(const T &t, const Args &... data)
84 {
85 _write(t, data...);
86 }
87 int a[maxn], vis[maxn];
88 int st[maxn];
89 int main(int argc, char const *argv[])
90 {
91 #ifndef ONLINE_JUDGE
92 freopen("in.txt", "r", stdin);
93 // freopen("out.txt","w", stdout);
94 #endif
95 n = read<int>(), m = read<int>(), k = read<int>();
96 int sum = 0;
97 for (int i = 1; i <= m; i++)
98 a[i] = read<int>(), sum += a[i];
99 int left = n - sum;
100 int mod = left % (m + 1);
101 int div = left / (m + 1);
102 int pdiv = div;
103 if (mod)
104 ++pdiv;
105 if (pdiv >=k||(k==1&&left))
106 puts("NO");
107 else
108 {
109 if (left <=m + 1)
110 {
111
112 int mod = left;
113 int now = 1;
114 for (int i = 1; i <= n;i++){
115 if (mod){
116 ++i;
117 --mod;
118 }
119 for (int j = i, p = 0; p < a[now]; j++, p++)
120 vis[j] = now;
121 i += a[now] - 1;
122 ++now;
123 }
124 }
125 else
126 {
127 int div = left / (m + 1);
128 int mod = left % (m + 1);
129 int now = 1;
130 for (int i = 1; i <= n;i++)
131 {
132 if (mod)
133 {
134 ++i;
135 --mod;
136 }
137 i += div;
138 for (int j = i, p = 0; p < a[now]; j++, p++)
139 vis[j] = now;
140 i += a[now] - 1;
141 ++now;
142 }
143 }
144 puts("YES");
145 for (int i = 1; i <= n; i++)
146 write(vis[i]), pblank;
147 puts("");
148 }
149 return 0;
150 }
D. Binary String Minimizing
Description
给出一个只包含01的字符串，可以进行k次操作。
每次操作可以将任意相邻两个值交换。
问最小的字典序。
Solution
讲道理这个题比BC简单好吧。
直接贪心，将0移到可能移到的最前面。
1 #include <algorithm>
2 #include <numeric>
3 #include <cctype>
4 #include <cmath>
5 #include <cstdio>
6 #include <cstdlib>
7 #include <cstring>
8 #include <iostream>
9 #include <map>
10 #include <queue>
11 #include <set>
12 #include <stack>
13 #if __cplusplus >= 201103L
14 #include <unordered_map>
15 #include <unordered_set>
16 #endif
17 #include <vector>
18 #define lson rt << 1, l, mid
19 #define rson rt << 1 | 1, mid + 1, r
20 #define LONG_LONG_MAX 9223372036854775807LL
21 #define pblank putchar(' ')
22 #define ll LL
23 #define fastIO ios::sync_with_stdio(false), cin.tie(0), cout.tie(0)
24 using namespace std;
25 typedef long long ll;
26 typedef long double ld;
27 typedef unsigned long long ull;
28 typedef pair<int, int> P;
29 ll n, m, k;
30 const int maxn = 1e6 + 10;
31 template <class T>
32 inline T read()
33 {
34 int f = 1;
35 T ret = 0;
36 char ch = getchar();
37 while (!isdigit(ch))
38 {
39 if (ch == '-')
40 f = -1;
41 ch = getchar();
42 }
43 while (isdigit(ch))
44 {
45 ret = (ret << 1) + (ret << 3) + ch - '0';
46 ch = getchar();
47 }
48 ret *= f;
49 return ret;
50 }
51 template <class T>
52 inline void write(T n)
53 {
54 if (n < 0)
55 {
56 putchar('-');
57 n = -n;
58 }
59 if (n >= 10)
60 {
61 write(n / 10);
62 }
63 putchar(n % 10 + '0');
64 }
65 template <class T>
66 inline void writeln(const T &n)
67 {
68 write(n);
69 puts("");
70 }
71 template <typename T>
72 void _write(const T &t)
73 {
74 write(t);
75 }
76 template <typename T, typename... Args>
77 void _write(const T &t, Args... args)
78 {
79 write(t), pblank;
80 _write(args...);
81 }
82 template <typename T, typename... Args>
83 inline void write_line(const T &t, const Args &... data)
84 {
85 _write(t, data...);
86 }
87 char s[maxn];
88 int lftz[maxn],lfto[maxn];
89 int main(int argc, char const *argv[])
90 {
91 #ifndef ONLINE_JUDGE
92 freopen("in.txt","r", stdin);
93 // freopen("out.txt","w", stdout);
94 #endif
95 fastIO;
96 int t;
97 cin >> t;
98 while(t--){
99 cin >> n >> k;
100 cin >> s+1;
101 for (int i = 1; i <= n;i++)
102 if (s[i]=='1')
103 lfto[i] = lfto[i - 1] + 1, lftz[i] = lftz[i - 1];
104 else
105 lftz[i] = lftz[i - 1] + 1, lfto[i] = lfto[i - 1];
106 for (int i = 1; i <= n&&k;i++)
107 if (s[i]=='0'){
108 if (lfto[i-1]<=k){
109 int pre = lftz[i - 1];
110 swap(s[pre + 1], s[i]);
111 k -=lfto[i - 1];
112 }
113 else{
114 swap(s[i], s[i-k]);
115 k = 0;
116 }
117 }
118 cout << s + 1 << "\n";
119 }
120 return 0;
121 }
F. Equalizing Two Strings
Description
给出两个字符串s,t，每次可以选择等长的两个不要求相同的字串进行翻转。
问能否将st翻转为相同。
Solution
1，字母数量不同$\rightarrow NO$
2，字母数量相同且存在大于1$\rightarrow YES$
3，字母相同且数目均为1,则考虑两个序列奇偶性是否相同。
1 #include <algorithm>
2 #include <numeric>
3 #include <cctype>
4 #include <cmath>
5 #include <cstdio>
6 #include <cstdlib>
7 #include <cstring>
8 #include <iostream>
9 #include <map>
10 #include <queue>
11 #include <set>
12 #include <stack>
13 #if __cplusplus >= 201103L
14 #include <unordered_map>
15 #include <unordered_set>
16 #endif
17 #include <vector>
18 #define lson rt << 1, l, mid
19 #define rson rt << 1 | 1, mid + 1, r
20 #define LONG_LONG_MAX 9223372036854775807LL
21 #define pblank putchar(' ')
22 #define ll LL
23 #define fastIO ios::sync_with_stdio(false), cin.tie(0), cout.tie(0)
24 using namespace std;
25 typedef long long ll;
26 typedef long double ld;
27 typedef unsigned long long ull;
28 typedef pair<int, int> P;
29 int n, m, k;
30 const int maxn = 2e5 + 10;
31 template <class T>
32 inline T read()
33 {
34 int f = 1;
35 T ret = 0;
36 char ch = getchar();
37 while (!isdigit(ch))
38 {
39 if (ch == '-')
40 f = -1;
41 ch = getchar();
42 }
43 while (isdigit(ch))
44 {
45 ret = (ret << 1) + (ret << 3) + ch - '0';
46 ch = getchar();
47 }
48 ret *= f;
49 return ret;
50 }
51 template <class T>
52 inline void write(T n)
53 {
54 if (n < 0)
55 {
56 putchar('-');
57 n = -n;
58 }
59 if (n >= 10)
60 {
61 write(n / 10);
62 }
63 putchar(n % 10 + '0');
64 }
65 template <class T>
66 inline void writeln(const T &n)
67 {
68 write(n);
69 puts("");
70 }
71 template <typename T>
72 void _write(const T &t)
73 {
74 write(t);
75 }
76 template <typename T, typename... Args>
77 void _write(const T &t, Args... args)
78 {
79 write(t), pblank;
80 _write(args...);
81 }
82 template <typename T, typename... Args>
83 inline void write_line(const T &t, const Args &... data)
84 {
85 _write(t, data...);
86 }
87 inline int lowbit(int x){
88 return x & (-x);
89 }
90 struct node{
91 char ch;
92 int id;
93 node(){}
94 node(char ch,int id){
95 this->ch = ch;
96 this->id = id;
97 }
98 };
99 char s1[maxn], s2[maxn];
100 int main(int argc, char const *argv[])
101 {
102 #ifndef ONLINE_JUDGE
103 freopen("in.txt","r", stdin);
104 // freopen("out.txt","w", stdout);
105 #endif
106 fastIO;
107 int t;
108 cin >> t;
109 while(t--){
110 cin >> n;
111 cin >> s1 >> s2;
112 vector<int> num1(26, 0), num2(26, 0);
113 for (int i = 0; i < n;i++)
114 num1[s1[i] - 'a']++, num2[s2[i] - 'a']++;
115 int f = 1;
116 for (int i = 0; i < 26;i++)
117 if (num1[i]!=num2[i]){
118 f = 0;
119 break;
120 }
121 else{
122 if (num1[i]>1)
123 f = 2;
124 }
125 if (!f)
126 puts("NO");
127 else if (f==2)
128 puts("YES");
129 else{
130 int f1 = 0, f2 = 0;
131 for (int i = 0; i < n;i++)
132 for (int j = 0; j < i;j++){
133 if (s1[i]>s1[j])
134 ++f1;
135 if (s2[i]>s2[j])
136 ++f2;
137 }
138 if ((f1+f2)%2==0)
139 puts("YES");
140 else
141 puts("NO");
142 }
143 }
144 return 0;
145 }