A Vitaly and Strings
题意：给定长度相等的字符串s，t，且s的字典序小于t，求一个字符串q字典序大于s小于t。
分析：将字符串看做26进制的数，对s”+1“即可。
#include <bits/stdc++.h>
#define pb push_back
#define mp make_pair
#define esp 1e-14
#define lson l, m, rt<<1
#define rson m+1, r, rt<<1|1
#define low(x) ((x)&(-(x)))
#define sz(x) ((int)((x).size()))
#define pf(x) ((x)*(x))
#define pb push_back
#define pi acos(-1.0)
#define in freopen("solve_in.txt", "r", stdin);
#define bug(x) printf("Line : %u >>>>>>\n", (x));
#define TL cerr << "Time elapsed: " << (double)clock() / CLOCKS_PER_SEC * 1000 << " ms" << endl;
#define inf 0x0f0f0f0f
using namespace std;
typedef long long LL;
typedef unsigned US;
typedef pair<int, int> PII;
typedef map<PII, int> MPS;
typedef MPS::iterator IT;
char sa[111], sb[111];
int main(){
// in
cin >> sa >> sb;
int n = strlen(sa);
int ok = 0;
for(int i = n-1; i >= 0; i--){
if(sa[i] != ‘z‘) {
sa[i] = sa[i] + 1;
ok = 1;
break;
}else sa[i] = ‘a‘;
}
if(ok) {
ok = strcmp(sa, sb) < 0;
}
if(ok) cout << sa << endl;
else puts("No such string");
return 0;
}
B Tanya and Postcard
题意：给定s和t，t的长度>=s，t中选出一些字符构成一个字符串，要求对应位置和s完全相同的尽量多，
然后剩下的要求对应位置仅仅大小写不同的尽量多。
分析：用t优先给s中字符找到完全相同的字符匹配，然后再利用大小写不同的字符匹配。
#include <bits/stdc++.h>
#define pb push_back
#define mp make_pair
#define esp 1e-14
#define lson l, m, rt<<1
#define rson m+1, r, rt<<1|1
#define low(x) ((x)&(-(x)))
#define sz(x) ((int)((x).size()))
#define pf(x) ((x)*(x))
#define pb push_back
#define pi acos(-1.0)
#define in freopen("solve_in.txt", "r", stdin);
#define bug(x) printf("Line : %u >>>>>>\n", (x));
#define TL cerr << "Time elapsed: " << (double)clock() / CLOCKS_PER_SEC * 1000 << " ms" << endl;
#define inf 0x0f0f0f0f
using namespace std;
typedef long long LL;
typedef unsigned US;
typedef pair<int, int> PII;
typedef map<PII, int> MPS;
typedef MPS::iterator IT;
const int maxn = (int)2e5 + 100;
char sa[maxn], sb[maxn];
int cntA[30][2], cntB[30][2];
int main(){
// in
int ans1 = 0, ans2 = 0;
scanf("%s%s", sa, sb);
int n = strlen(sa);
int m = strlen(sb);
for(int i = 0; i < n; i++){
if(sa[i] >= ‘a‘ && sa[i] <= ‘z‘){
cntA[sa[i]-‘a‘][0]++;
}else{
cntA[sa[i]-‘A‘][1]++;
}
}
for(int i = 0; i < m; i++){
if(sb[i] >= ‘a‘ && sb[i] <= ‘z‘){
cntB[sb[i]-‘a‘][0]++;
}else{
cntB[sb[i]-‘A‘][1]++;
}
}
for(int i = 0; i < 26; i++){
int tmp1, tmp2;
ans1 += (tmp1 = min(cntA[i][0], cntB[i][0])) + (tmp2 = min(cntA[i][1], cntB[i][1]));
cntA[i][0] -= tmp1;
cntB[i][0] -= tmp1;
cntA[i][1] -= tmp2;
cntB[i][1] -= tmp2;
ans2 += min(cntA[i][1], cntB[i][0]);
ans2 += min(cntA[i][0], cntB[i][1]);
}
printf("%d %d\n", ans1, ans2);
return 0;
}
C Anya and Smartphone
题意：n个程序，每个屏幕上最多k个，给定放置顺序，以及依次运行的m个程序，每次运行后位置向前移动一位，
位于第i个屏幕的程序需要i此操作。
分析：直接模拟每次操作，运行后交换一下和之前的程序的位置。
#include <bits/stdc++.h>
#define pb push_back
#define mp make_pair
#define esp 1e-14
#define lson l, m, rt<<1
#define rson m+1, r, rt<<1|1
#define low(x) ((x)&(-(x)))
#define sz(x) ((int)((x).size()))
#define pf(x) ((x)*(x))
#define pb push_back
#define pi acos(-1.0)
#define in freopen("solve_in.txt", "r", stdin);
#define bug(x) printf("Line : %u >>>>>>\n", (x));
#define TL cerr << "Time elapsed: " << (double)clock() / CLOCKS_PER_SEC * 1000 << " ms" << endl;
#define inf 0x0f0f0f0f
using namespace std;
typedef long long LL;
typedef unsigned US;
typedef pair<int, int> PII;
typedef map<PII, int> MPS;
typedef MPS::iterator IT;
const int maxn = (int)1e5 + 10;
int a[maxn], pos[maxn];
int main(){
// in
int n, m, k;
LL ans = 0;
cin >> n >> m >> k;
for(int i = 1; i <= n; i++){
scanf("%d", a+i);
pos[a[i]] = i;
}
for(int i = 1; i <= m; i++){
int u;
scanf("%d", &u);
ans += pos[u]/k + (pos[u]%k != 0);
if(pos[u] != 1) {
swap(a[pos[u]-1], a[pos[u]]);
int t = pos[u];
pos[a[t]] = t;
pos[a[t-1]] = t-1;
}
}
printf("%I64d\n", ans);
return 0;
}
D Ilya and Escalator
题意：n个人排队依次上电梯，每个人进入或继续等待电梯的概率分别为p，1-p，每秒钟最多只有队首的人进入电梯，
求t秒后在电梯上的人数期望值。
分析：dp[t][m] 表示t秒时有m个人上电梯的概率，转移：dp[t][m] = dp[t-1][m]*(1-p)+dp[t-1][m-1]*p，注意一下边界。
#include <bits/stdc++.h>
#define pb push_back
#define mp make_pair
#define esp 1e-14
#define lson l, m, rt<<1
#define rson m+1, r, rt<<1|1
#define low(x) ((x)&(-(x)))
#define sz(x) ((int)((x).size()))
#define pf(x) ((x)*(x))
#define pb push_back
#define pi acos(-1.0)
#define in freopen("solve_in.txt", "r", stdin);
#define bug(x) printf("Line : %u >>>>>>\n", (x));
#define TL cerr << "Time elapsed: " << (double)clock() / CLOCKS_PER_SEC * 1000 << " ms" << endl;
#define inf 0x0f0f0f0f
using namespace std;
typedef long long LL;
typedef unsigned US;
typedef pair<int, int> PII;
typedef map<PII, int> MPS;
typedef MPS::iterator IT;
const int maxn = 2000 + 10;
double dp[maxn][maxn];
int main() {
// in
int n, m;
double p;
scanf("%d%lf%d", &n, &p, &m);
dp[0][0] = 1.0;
for(int i = 1; i <= m; i++)
for(int j = 0; j <= min(i, n); j++) {
if(j != 0)
dp[i][j] += dp[i-1][j-1]*p;
if(j == n)
dp[i][j] += dp[i-1][j];
else
dp[i][j] += dp[i-1][j]*(1-p);
}
double ans = 0;
for(int i = 0; i <= n; i++)
ans += dp[m][i]*i;
printf("%.12f\n", ans);
return 0;
}
E Arthur and Questions
题意：给定数组a[]，长度n <= 1e5，以及k，要求满足 (a1 + a2 ... + ak, a2 + a3 + ... + ak + 1, ..., an - k + 1 + an - k + 2 + ... + an)构成递增
序列，其中有些ai是未知的，要求满足条件的a[i]序列且|a[i]的和尽量小。
分析：由上面序列递增容易得到:
a1 < ak+1 < a2k+1 < ... < apk+1
a2 < ak+2 < a2k+2 < ... < aqk+1
....
ak < a2K < a3k < ... < ark
每个不等式组独立，以a1 < ak+1 < a2k+1 < ... < apk+1为例，考虑其中两个值确定(不为?)的aik+1及ajk+1，
之间的alk+1的取值未定，那么alk+1取值肯定在aik+1~ajk+1之间，记做[l, r]，数的个数必须满足r-l+1>=j-i-1，记num = j-i-1。
分为以下几种情况：
r <= 0 : 将alk+1取r-num+1, r-num+2, ....r即可；
l >= 0 : 将alk+1取l, l+1, l+2,........l+num-1即可；
l <0<r :按照0,±1,±2,±3.....±n等，注意边界范围不足时进行适当处理。
对于仍然为？的a[i]取0即可。
代码：
1 #include <bits/stdc++.h>
2 #define pb push_back
3 #define mp make_pair
4 #define esp 1e-14
5 #define lson l, m, rt<<1
6 #define rson m+1, r, rt<<1|1
7 #define low(x) ((x)&(-(x)))
8 #define sz(x) ((int)((x).size()))
9 #define pf(x) ((x)*(x))
10 #define pb push_back
11 #define pi acos(-1.0)
12 #define in freopen("solve_in.txt", "r", stdin);
13 #define bug(x) printf("Line : %u >>>>>>\n", (x));
14 #define TL cerr << "Time elapsed: " << (double)clock() / CLOCKS_PER_SEC * 1000 << " ms" << endl;
15 #define inf 0x0f0f0f0f
16
17
18 using namespace std;
19 typedef long long LL;
20 typedef unsigned US;
21 typedef pair<int, int> PII;
22 typedef map<PII, int> MPS;
23 typedef MPS::iterator IT;
24 const int maxn = (int)1e5 + 10;
25 const int INF = (int)2e9;
26 int a[maxn];
27
28 int main() {
29 // in
30 int n, k;
31 cin >> n >> k;
32 getchar();
33 for(int i = 1; i <= n; i++) {
34 char ch;
35 while((ch = getchar()) == ‘ ‘);
36 if(isdigit(ch) || ch == ‘-‘) {
37 ungetc(ch, stdin);
38 scanf("%d", a+i);
39 } else a[i] = INF;
40 }
41 a[0] = -(int)1e9 - maxn;
42 a[n+1] = (int)1e9 + maxn;
43 int ok = 1;
44 for(int st = 1; st <= k && ok; st++) {
45 if(st+k > n) break;
46 vector<int> index;
47 index.pb(0);
48 for(int now = st; now <= n; now += k)
49 index.pb(now);
50 index.pb(n+1);
51 int first = 0, next = 1;
52
53 while(next < sz(index) && ok) {
54 while(a[index[next]] == INF)
55 next++;
56
57 int l = a[index[first]];
58 int r = a[index[next]];
59 if(r-l-1 < next-first-1) {
60 ok = 0;
61 break;
62 }
63 l++, r--;
64 vector<int> val;
65 int num = next-first-1;
66 if(l >= 0) {
67 while(num--) {
68 val.pb(l+num);
69 }
70 } else if(r <= 0) {
71 while(num--) {
72 val.pb(r-num);
73 }
74 } else {
75 int v = 0;
76 for(int p = 0; p < num; p++) {
77 int parity = p&1;
78 if(p) {
79 if(parity) {
80 v = -v+1;
81 if(v > r) {
82 v = -v;
83 while(p < num) {
84 val.pb(v);
85 v--;
86 p++;
87 }
88 break;
89 }
90 } else {
91 v = -v;
92 if(v < l) {
93 v = -v+1;
94 while(p < num) {
95 val.pb(v);
96 v++;
97 p++;
98 }
99 break;
100 }
101 }
102 }
103 val.pb(v);
104 }
105 }
106
107 sort(val.begin(), val.end());
108 int cnt = 0;
109
110 while(++first < next) {
111 a[index[first]] = val[cnt++];
112 }
113 next++;
114 }
115 }
116 if(ok) {
117 for(int i = 1; i <= n; i++)
118 printf("%d%c", a[i] == INF ? 0 : a[i], i == n ? ‘\n‘ : ‘ ‘);
119 } else puts("Incorrect sequence");
120
121 return 0;
122 }
F Pasha and Pipe
代码：
时间： 2024-10-25 18:06:30