[Educational Codeforces Round 39 (Rated for Div. 2)] 题解
A. Partition
思路：
贪心
让正数都放入集合(mathit B)，非正数都放入集合$mathit C $即可。
代码：
int n;
ll a[maxn];
int main()
{
#if DEBUG_Switch
freopen("C:\code\input.txt", "r", stdin);
#endif
//freopen("C:\code\output.txt","w",stdout);
cin >> n;
ll ans = 0ll;
repd(i, 1, n)
{
cin >> a[i];
ans += abs(a[i]);
}
cout << ans << endl;
return 0;
}
B. Weird Subtraction Process
思路：
直接按照操作模拟，不过要把减法变成取模，这样并不影响结果，变成取模运算之后，就可以(O(logn))的时间复杂度结束操纵了。
代码：
/*** TEMPLATE CODE * * STARTS HERE ***/
#define DEBUG_Switch 0
int main()
{
#if DEBUG_Switch
freopen("C:\code\input.txt", "r", stdin);
#endif
//freopen("C:\code\output.txt","w",stdout);
ll a, b;
cin >> a >> b;
while (1)
{
if (a == 0 || b == 0)
break;
if (a >= b * 2)
{
a %= b * 2;
continue;
} else if (b >= a * 2)
{
b %= a * 2;
continue;
} else
{
break;
}
}
cout << a << " " << b << endl;
return 0;
}
C. String Transformation
思路：
算法：贪心
先把目标字符设为‘a’，找到第一个字典序小于等于字符‘a'的，将其变成’a'，
然后把目标字符变成‘b’，找到第一个字典序小于等于字符‘b'的，将其变成’b'，
然后把目标字符变成‘c’，以此下去，直到除了完了‘z’字符。
如果无法处理到‘z’字符，就输出-1.
代码：
/*** TEMPLATE CODE * * STARTS HERE ***/
#define DEBUG_Switch 0
char s[maxn];
int n;
int main()
{
#if DEBUG_Switch
freopen("C:\code\input.txt", "r", stdin);
#endif
//freopen("C:\code\output.txt","w",stdout);
scanf("%s", s + 1);
n = strlen(s + 1);
char x = 'a';
int flag = 0;
repd(i, 1, n)
{
if (s[i] <= x)
{
s[i] = x;
if (x < 'z')
x++;
else
flag = 1;
}
}
if (flag)
printf("%s
", s + 1 );
else
printf("-1
");
return 0;
}
D. Timetable
思路：
可以很简单的处理出：(info[i][j])代表第(mathit i)天逃了(mathit j)个课程，最多可以少去学校多久。
然后分组背包求(dp_i)代表一共逃了i节课最多可以少去学校多久。
具体是：对于第(mathit i)天，枚举逃了(mathit j)个课程，此时会产生一个体积和一个价值，然后去DP背包即可。
代码：
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <bits/stdc++.h>
#define ALL(x) (x).begin(), (x).end()
#define sz(a) int(a.size())
#define rep(i,x,n) for(int i=x;i<n;i++)
#define repd(i,x,n) for(int i=x;i<=n;i++)
#define pii pair<int,int>
#define pll pair<long long ,long long>
#define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define MS0(X) memset((X), 0, sizeof((X)))
#define MSC0(X) memset((X), ' ', sizeof((X)))
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define eps 1e-6
#define chu(x) if(DEBUG_Switch) cout<<"["<<#x<<" "<<(x)<<"]"<<endl
#define du3(a,b,c) scanf("%d %d %d",&(a),&(b),&(c))
#define du2(a,b) scanf("%d %d",&(a),&(b))
#define du1(a) scanf("%d",&(a));
using namespace std;
typedef long long ll;
ll gcd(ll a, ll b) {return b ? gcd(b, a % b) : a;}
ll lcm(ll a, ll b) {return a / gcd(a, b) * b;}
ll powmod(ll a, ll b, ll MOD) { if (a == 0ll) {return 0ll;} a %= MOD; ll ans = 1; while (b) {if (b & 1) {ans = ans * a % MOD;} a = a * a % MOD; b >>= 1;} return ans;}
ll poww(ll a, ll b) { if (a == 0ll) {return 0ll;} ll ans = 1; while (b) {if (b & 1) {ans = ans * a ;} a = a * a ; b >>= 1;} return ans;}
void Pv(const vector<int> &V) {int Len = sz(V); for (int i = 0; i < Len; ++i) {printf("%d", V[i] ); if (i != Len - 1) {printf(" ");} else {printf("
");}}}
void Pvl(const vector<ll> &V) {int Len = sz(V); for (int i = 0; i < Len; ++i) {printf("%lld", V[i] ); if (i != Len - 1) {printf(" ");} else {printf("
");}}}
inline long long readll() {long long tmp = 0, fh = 1; char c = getchar(); while (c < '0' || c > '9') {if (c == '-') fh = -1; c = getchar();} while (c >= '0' && c <= '9') tmp = tmp * 10 + c - 48, c = getchar(); return tmp * fh;}
inline int readint() {int tmp = 0, fh = 1; char c = getchar(); while (c < '0' || c > '9') {if (c == '-') fh = -1; c = getchar();} while (c >= '0' && c <= '9') tmp = tmp * 10 + c - 48, c = getchar(); return tmp * fh;}
void pvarr_int(int *arr, int n, int strat = 1) {if (strat == 0) {n--;} repd(i, strat, n) {printf("%d%c", arr[i], i == n ? '
' : ' ');}}
void pvarr_LL(ll *arr, int n, int strat = 1) {if (strat == 0) {n--;} repd(i, strat, n) {printf("%lld%c", arr[i], i == n ? '
' : ' ');}}
const int maxn = 510;
const int inf = 0x3f3f3f3f;
/*** TEMPLATE CODE * * STARTS HERE ***/
#define DEBUG_Switch 0
int n, m, k;
char s[maxn][maxn];
int sum[maxn][maxn];
int info[maxn][maxn];
int dp[maxn];
int f[maxn];
int len[maxn];
int main()
{
#if DEBUG_Switch
freopen("C:\code\input.txt", "r", stdin);
#endif
//freopen("C:\code\output.txt","w",stdout);
n = readint();
m = readint();
k = readint();
int ans = 0;
repd(i, 1, n)
{
scanf("%s", s[i] + 1);
int pre = 0;
int last = 0;
repd(j, 1, m)
{
if (s[i][j] == '1')
{
if (pre == 0)
pre = j;
last = j;
sum[i][j] = sum[i][j - 1] + 1;
} else
sum[i][j] = sum[i][j - 1];
}
if (last != 0) {
len[i] = last - pre + 1;
ans += last - pre + 1;
} else
{
len[i] = 0;
}
}
repd(i, 1, n)
{
repd(j, 1, m)
{
if (s[i][j] != '1')
continue;
int tot = sum[i][m];
info[i][tot] = max(info[i][tot], len[i]);
repd(w, j, m)
{
if (s[i][w] != '1')
continue;
int have = sum[i][w] - sum[i][j - 1];
int skip = tot - have;
info[i][skip] = max(info[i][skip], len[i] - (w - j + 1));
}
}
}
// repd(i, 1, n)
// {
// repd(j, 0, m)
// {
// cout << i << " " << j << " " << info[i][j] << endl;
// }
// }
repd(i, 1, n)
{
repd(w, 1, k)
{
int val = info[i][w];
for (int j = k; j >= w; --j)
{
f[j] = max(f[j], dp[j - w] + val);
}
}
repd(j, 1, k)
{
dp[j] = max(dp[j], f[j]);
f[j] = 0;
}
}
int num = 0;
repd(i, 1, k)
{
num = max(num, dp[i]);
}
ans -= num;
printf("%d
", ans );
return 0;
}
E. Largest Beautiful Number
思路：
贪心，实现
贪心思想：让答案字符串和给定字符串的最大公共前缀（LCP）尽可能长。
我们通过dfs枚举LCP的长度，然后判断是否如何条件即可。
注意，如果需要高位降低才能符合条件，那么要在中间加一段‘9’,尾部以降序输出当前仍是出现奇数次数的数字。
代码：
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <bits/stdc++.h>
#define ALL(x) (x).begin(), (x).end()
#define sz(a) int(a.size())
#define rep(i,x,n) for(int i=x;i<n;i++)
#define repd(i,x,n) for(int i=x;i<=n;i++)
#define pii pair<int,int>
#define pll pair<long long ,long long>
#define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define MS0(X) memset((X), 0, sizeof((X)))
#define MSC0(X) memset((X), ' ', sizeof((X)))
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define eps 1e-6
#define chu(x) if(DEBUG_Switch) cout<<"["<<#x<<" "<<(x)<<"]"<<endl
#define du3(a,b,c) scanf("%d %d %d",&(a),&(b),&(c))
#define du2(a,b) scanf("%d %d",&(a),&(b))
#define du1(a) scanf("%d",&(a));
using namespace std;
typedef long long ll;
ll gcd(ll a, ll b) {return b ? gcd(b, a % b) : a;}
ll lcm(ll a, ll b) {return a / gcd(a, b) * b;}
ll powmod(ll a, ll b, ll MOD) { if (a == 0ll) {return 0ll;} a %= MOD; ll ans = 1; while (b) {if (b & 1) {ans = ans * a % MOD;} a = a * a % MOD; b >>= 1;} return ans;}
ll poww(ll a, ll b) { if (a == 0ll) {return 0ll;} ll ans = 1; while (b) {if (b & 1) {ans = ans * a ;} a = a * a ; b >>= 1;} return ans;}
void Pv(const vector<int> &V) {int Len = sz(V); for (int i = 0; i < Len; ++i) {printf("%d", V[i] ); if (i != Len - 1) {printf(" ");} else {printf("
");}}}
void Pvl(const vector<ll> &V) {int Len = sz(V); for (int i = 0; i < Len; ++i) {printf("%lld", V[i] ); if (i != Len - 1) {printf(" ");} else {printf("
");}}}
inline long long readll() {long long tmp = 0, fh = 1; char c = getchar(); while (c < '0' || c > '9') {if (c == '-') fh = -1; c = getchar();} while (c >= '0' && c <= '9') tmp = tmp * 10 + c - 48, c = getchar(); return tmp * fh;}
inline int readint() {int tmp = 0, fh = 1; char c = getchar(); while (c < '0' || c > '9') {if (c == '-') fh = -1; c = getchar();} while (c >= '0' && c <= '9') tmp = tmp * 10 + c - 48, c = getchar(); return tmp * fh;}
void pvarr_int(int *arr, int n, int strat = 1) {if (strat == 0) {n--;} repd(i, strat, n) {printf("%d%c", arr[i], i == n ? '
' : ' ');}}
void pvarr_LL(ll *arr, int n, int strat = 1) {if (strat == 0) {n--;} repd(i, strat, n) {printf("%lld%c", arr[i], i == n ? '
' : ' ');}}
const int maxn = 1000010;
const int inf = 0x3f3f3f3f;
/*** TEMPLATE CODE * * STARTS HERE ***/
#define DEBUG_Switch 0
char s[maxn];
int n;
bool solve(int id, int info, int last, int eq)
{
if (!eq)
{
if (id + __builtin_popcount(info) <= n)
{
repd(i, 0, id - 2)
{
printf("%c", s[i]);
}
printf("%d", last);
int rm = n - id - __builtin_popcount(info);
while (rm--)
{
printf("9");
}
for (int i = 9; i >= 0; --i)
{
if ((1 << i)&info)
{
printf("%d", i );
}
}
printf("
");
return 1;
}
return 0;
}
for (int i = s[id] - '0'; i >= 0 + (!id); --i)
{
if (solve(id + 1, info ^ (1 << i), i, eq & (i == s[id] - '0')))
{
return 1;
}
}
return 0;
}
int main()
{
#if DEBUG_Switch
freopen("C:\code\input.txt", "r", stdin);
#endif
//freopen("C:\code\output.txt","w",stdout);
int t;
t = readint();
while (t--)
{
scanf("%s", s);
n = strlen(s);
if (!solve(0, 0, -1, 1))
{
repd(i, 1, n - 2)
{
printf("9");
}
printf("
");
}
}
return 0;
}
F. Fibonacci String Subsequences
思路：
动态规划
状态：
(dp[i][l][r])为(F(i))的所有子串中(s_{lsim r})出现的次数。
考虑转移：一共有三部分：
1、(s_{lsim r})都在(F(i-1))中，如果(r=n)，那么(F(i-2))中的字符可以随机选择，所以有(2^{len(F(i-2))})种，若(r ot=n),则不能在后面乱选，所以贡献只有1种。
2、(s_{lsim r})都在(F(i-2))中，如果(l=1)，那么(F(i-1))中的字符可以随机选择，所以有(2^{len(F(i-1))})种，若(l ot=1),则不能在前面乱选，所以贡献只有1种。
3、(s_{lsim r})都分在(F(i-1),F(i-2))中，我们只需要枚举(kin[l,r-1])作为分界点，(s_{l,k})在(F(i-1))中，(s_{k+1,r})在(F(i-2))中，则贡献为(dp[i-1][l][k]*dp[i-2][k+1][r])种。
代码：
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <bits/stdc++.h>
#define ALL(x) (x).begin(), (x).end()
#define sz(a) int(a.size())
#define rep(i,x,n) for(int i=x;i<n;i++)
#define repd(i,x,n) for(int i=x;i<=n;i++)
#define pii pair<int,int>
#define pll pair<long long ,long long>
#define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define MS0(X) memset((X), 0, sizeof((X)))
#define MSC0(X) memset((X), ' ', sizeof((X)))
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define eps 1e-6
#define chu(x) if(DEBUG_Switch) cout<<"["<<#x<<" "<<(x)<<"]"<<endl
#define du3(a,b,c) scanf("%d %d %d",&(a),&(b),&(c))
#define du2(a,b) scanf("%d %d",&(a),&(b))
#define du1(a) scanf("%d",&(a));
using namespace std;
typedef long long ll;
ll gcd(ll a, ll b) {return b ? gcd(b, a % b) : a;}
ll lcm(ll a, ll b) {return a / gcd(a, b) * b;}
ll powmod(ll a, ll b, ll MOD) { if (a == 0ll) {return 0ll;} a %= MOD; ll ans = 1; while (b) {if (b & 1) {ans = ans * a % MOD;} a = a * a % MOD; b >>= 1;} return ans;}
ll poww(ll a, ll b) { if (a == 0ll) {return 0ll;} ll ans = 1; while (b) {if (b & 1) {ans = ans * a ;} a = a * a ; b >>= 1;} return ans;}
void Pv(const vector<int> &V) {int Len = sz(V); for (int i = 0; i < Len; ++i) {printf("%d", V[i] ); if (i != Len - 1) {printf(" ");} else {printf("
");}}}
void Pvl(const vector<ll> &V) {int Len = sz(V); for (int i = 0; i < Len; ++i) {printf("%lld", V[i] ); if (i != Len - 1) {printf(" ");} else {printf("
");}}}
inline long long readll() {long long tmp = 0, fh = 1; char c = getchar(); while (c < '0' || c > '9') {if (c == '-') fh = -1; c = getchar();} while (c >= '0' && c <= '9') tmp = tmp * 10 + c - 48, c = getchar(); return tmp * fh;}
inline int readint() {int tmp = 0, fh = 1; char c = getchar(); while (c < '0' || c > '9') {if (c == '-') fh = -1; c = getchar();} while (c >= '0' && c <= '9') tmp = tmp * 10 + c - 48, c = getchar(); return tmp * fh;}
void pvarr_int(int *arr, int n, int strat = 1) {if (strat == 0) {n--;} repd(i, strat, n) {printf("%d%c", arr[i], i == n ? '
' : ' ');}}
void pvarr_LL(ll *arr, int n, int strat = 1) {if (strat == 0) {n--;} repd(i, strat, n) {printf("%lld%c", arr[i], i == n ? '
' : ' ');}}
const int maxn = 102;
const int inf = 0x3f3f3f3f;
/*** TEMPLATE CODE * * STARTS HERE ***/
#define DEBUG_Switch 0
int n;
int x;
const ll mod = 1e9 + 7;
ll dp[maxn][maxn][maxn];
char s[maxn];
ll pow2[maxn];
int main()
{
#if DEBUG_Switch
freopen("C:\code\input.txt", "r", stdin);
#endif
//freopen("C:\code\output.txt","w",stdout);
n = readint();
x = readint();
scanf("%s", s + 1);
repd(i, 1, n)
{
dp[s[i] - '0'][i][i] = 1;
}
pow2[0] = pow2[1] = 2ll;
repd(i, 2, 100)
{
pow2[i] = pow2[i - 1] * pow2[i - 2] % mod;
}
repd(i, 2, x)
{
repd(l, 1, n)
{
repd(r, l, n)
{
dp[i][l][r] += dp[i - 1][l][r] * (r == n ? pow2[i - 2] : 1ll) % mod;
dp[i][l][r] %= mod;
dp[i][l][r] += dp[i - 2][l][r] * (l == 1 ? pow2[i - 1] : 1ll) % mod;
dp[i][l][r] %= mod;
for (int k = l; k < r; ++k)
{
dp[i][l][r] += dp[i - 1][l][k] * dp[i - 2][k + 1][r] % mod;
dp[i][l][r] %= mod;
}
}
}
}
printf("%lld
", dp[x][1][n]);
return 0;
}