SCAU2021春季个人排位赛第七场 （部分题解））
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A：折半搜索+二分 跟上星期一样的知识点
B：拓扑排序
C：里面知识点都经常考并且糅合在一起，非常好的一道题。并查集+树DP考虑边的贡献
D：扫描线 上星期知识点 出这道题是因为可以用bitset暴力卡，可以学习bitset，但是记得补正解
E：思维题 拿来签到
F：弗洛伊德
G：高斯消元比较模板的题，但是很恶心，卡精度，卡eps不能调太大，而且卡long double
A题
CodeForces - 912E
Opposite to Grisha's nice behavior, Oleg, though he has an entire year at his disposal, didn't manage to learn how to solve number theory problems in the past year. That's why instead of Ded Moroz he was visited by his teammate Andrew, who solemnly presented him with a set of n distinct prime numbers alongside with a simple task: Oleg is to find the k-th smallest integer, such that all its prime divisors are in this set.
Input
The first line contains a single integer n (1 ≤ n ≤ 16).
The next line lists n distinct prime numbers p1, p2, ..., pn (2 ≤ pi ≤ 100) in ascending order.
The last line gives a single integer k (1 ≤ k). It is guaranteed that the k-th smallest integer such that all its prime divisors are in this set does not exceed 1018.
Output
Print a single line featuring the k-th smallest integer. It's guaranteed that the answer doesn't exceed 1018.
Examples
Input
3 2 3 5 7
Output
8
Input
5 3 7 11 13 31 17
Output
93
Note
The list of numbers with all prime divisors inside {2, 3, 5} begins as follows:
(1, 2, 3, 4, 5, 6, 8, ...)
The seventh number in this list (1-indexed) is eight.
知识点和上星期一样，折半搜索＋二分查找。
主要是一开始读不懂题，根本不知道是什么意思，试了好久才知道是包括1且因数要包括集合内的元素。
先用DFS枚举好所有状态的结果，然后两边双指针同时进行，二分枚举目前的mid是排在第几，直到找到ans为止。
代码：
#include <iostream>
#include <stdio.h>
#include <algorithm>
#include <stdlib.h>
#include <cstdio>
#include <cstdlib>
#include <string>
#include <cstring>
#include <string.h>
#include <math.h>
#include <cmath>
#include <stack>
#include <queue>
#include <list>
#include <vector>
using namespace std;
const long long oo=1e18;
int k;
int n;
long long p[20];
vector<long long> a[2];
void ready()
{
ios::sync_with_stdio(false),cin.tie(0);
cin>>n;
for(int i=1;i<=n;i++)
cin>>p[i];
cin>>k;
}
void dfs(int temp,int u,long long now)
{
a[temp].push_back(now);
for(int i=u;i<=n;i+=2)
if(oo/p[i]>=now)
dfs(temp,i,now*p[i]);
}
bool check(long long mid)
{
long long ki=0;
for(int i=(int)a[0].size()-1,j=0;i>=0;i--)
{
while(j<(int)a[1].size() && a[1][j]<=mid/a[0][i])
j++;
ki+=j;
}
return ki>=k;
}
void work()
{
dfs(0,1,1);
dfs(1,2,1);
sort(a[0].begin(),a[0].end());
sort(a[1].begin(),a[1].end());
long long l=1,r=oo,mid=(l+r)/2;
while(l<=r)
{
mid=(l+r)/2;
if(check(mid))
r=mid-1;
else
l=mid+1;
}
cout<<l;
return ;
}
int main()
{
ready();
work();
return 0;
}
B题
CodeForces - 770C
Now you can take online courses in the Berland State University! Polycarp needs to pass k main online courses of his specialty to get a diploma. In total n courses are availiable for the passage.
The situation is complicated by the dependence of online courses, for each course there is a list of those that must be passed before starting this online course (the list can be empty, it means that there is no limitation).
Help Polycarp to pass the least number of courses in total to get the specialty (it means to pass all main and necessary courses). Write a program which prints the order of courses.
Polycarp passes courses consistently, he starts the next course when he finishes the previous one. Each course can't be passed more than once.
Input
The first line contains n and k (1 ≤ k ≤ n ≤ 105) — the number of online-courses and the number of main courses of Polycarp's specialty.
The second line contains k distinct integers from 1 to n — numbers of main online-courses of Polycarp's specialty.
Then n lines follow, each of them describes the next course: the i-th of them corresponds to the course i. Each line starts from the integer ti (0 ≤ ti ≤ n - 1) — the number of courses on which the i-th depends. Then there follows the sequence of ti distinct integers from 1 to n — numbers of courses in random order, on which the i-th depends. It is guaranteed that no course can depend on itself.
It is guaranteed that the sum of all values ti doesn't exceed 105.
Output
Print -1, if there is no the way to get a specialty.
Otherwise, in the first line print the integer m — the minimum number of online-courses which it is necessary to pass to get a specialty. In the second line print m distinct integers — numbers of courses which it is necessary to pass in the chronological order of their passage. If there are several answers it is allowed to print any of them.
Examples
Input
6 2 5 3 0 0 0 2 2 1 1 4 1 5
Output
5 1 2 3 4 5
Input
9 3 3 9 5 0 0 3 9 4 5 0 0 1 8 1 6 1 2 2 1 2
Output
6 1 2 9 4 5 3
Input
3 3 1 2 3 1 2 1 3 1 1
Output
-1
Note
In the first test firstly you can take courses number 1 and 2, after that you can take the course number 4, then you can take the course number 5, which is the main. After that you have to take only the course number 3, which is the last not passed main course.
虽说是拓扑排序，但我感觉更像是思维DFS，主要在记录状态的时候注意一下。好吧就是记录状态这里要做好就行了，就是个普普通通的DFS。
保存状态和最后存进ans在DFS最后储存，这样能保证先储存的就是最里面的点。这个地方卡了我好久，比赛的时候就是卡在这个地方了。
注意要反方向构图！
代码：
#include <iostream>
#include <stdio.h>
#include <algorithm>
#include <stdlib.h>
#include <cstdio>
#include <cstdlib>
#include <string>
#include <cstring>
#include <string.h>
#include <math.h>
#include <cmath>
#include <stack>
#include <queue>
#include <list>
#include <vector>
using namespace std;
int n,k;
int p[100005],nex[100005],to[100005],pi;
int ki[100005];
int ans[100005],ansi;
bool q[100005];
int f[100005];
int cnt[100005];
bool ANS;
struct B{
int id;
int in_;
}t[100005];
void read_in(int u,int v)
{
pi++;nex[pi]=p[u];p[u]=pi;to[pi]=v;
}
bool cmp(B i,B j)
{
return i.in_<j.in_;
}
void ready()
{
ios::sync_with_stdio(false),cin.tie(0);
cin>>n>>k;
for(int i=1;i<=k;i++)
cin>>ki[i];
for(int i=1;i<=n;i++)
{
int ti;
cin>>ti;
while(ti--)
{
int vi;
cin>>vi;
read_in(i,vi);
}
}
}
void dfs(int u)
{
f[u]=-1;
for(int i=p[u],v=to[i];i;i=nex[i],v=to[i])
{
if(f[v]==-1)
{
ANS=true;
return ;
}
if(!f[v])
dfs(v);
}
f[u]=1;
ans[++ansi]=u;
}
void work()
{
for(int i=1;i<=k;i++)
{
if(!f[ki[i]])
dfs(ki[i]);
if(ANS)
{
cout<<-1;
return ;
}
}
cout<<ansi<<'\\n';
for(int i=1;i<=ansi;i++)
cout<<ans[i]<<' ';
return ;
}
int main()
{
ready();
work();
return 0;
}
/*
5 3
2 1 3
0
1 1
1 2
0
0
*/
E题
CodeForces - 1186C
Vus the Cossack has two binary strings, that is, strings that consist only of "0" and "1". We call these strings aa and bb. It is known that |b|≤|a||b|≤|a|, that is, the length of bb is at most the length of aa.
The Cossack considers every substring of length |b||b| in string aa. Let's call this substring cc. He matches the corresponding characters in bb and cc, after which he counts the number of positions where the two strings are different. We call this function f(b,c)f(b,c).
For example, let b=00110b=00110, and c=11000c=11000. In these strings, the first, second, third and fourth positions are different.
Vus the Cossack counts the number of such substrings cc such that f(b,c)f(b,c) is even.
For example, let a=01100010a=01100010 and b=00110b=00110. aa has four substrings of the length |b||b|: 0110001100, 1100011000, 1000110001, 0001000010.
- f(00110,01100)=2f(00110,01100)=2;
- f(00110,11000)=4f(00110,11000)=4;
- f(00110,10001)=4f(00110,10001)=4;
- f(00110,00010)=1f(00110,00010)=1.
Since in three substrings, f(b,c)f(b,c) is even, the answer is 33.
Vus can not find the answer for big strings. That is why he is asking you to help him.
Input
The first line contains a binary string aa (1≤|a|≤1061≤|a|≤106) — the first string.
The second line contains a binary string bb (1≤|b|≤|a|1≤|b|≤|a|) — the second string.
Output
Print one number — the answer.
Examples
Input
01100010 00110
Output
3
Input
1010111110 0110
Output
4
Note
The first example is explained in the legend.
In the second example, there are five substrings that satisfy us: 10101010, 01010101, 11111111, 11111111.
思维题。吃在了看不懂题的亏。不知道even是偶数的意思，也没去猜f（b，c）到底是什么意思。
题意也就是两串01组成的字符串a，b。|a| >= |b| ，在a中取 |b| 长度，和b异或后如果1的个数为偶数则这是一个目标子串。问a中有多少个长度为|b|且满足条件的子串。
思维题，找规律。7爷比完之后立刻和我说这肯定cf的题目。
如果两个子串中，1的个数同为偶数或者同为奇数，则不管1在不在同一个位置，异或之后剩下的1肯定都是偶数。所以用前缀和维护就好。
代码
#include <iostream>
#include <stdio.h>
#include <algorithm>
#include <stdlib.h>
#include <cstdio>
#include <cstdlib>
#include <string>
#include <cstring>
#include <string.h>
#include <math.h>
#include <cmath>
#include <stack>
#include <queue>
#include <list>
#include <vector>
using namespace std;
string a,b;
int la,lb,ai[1000006],bi[1000006],ans;
void ready()
{
ios::sync_with_stdio(false),cin.tie(0);
cin>>a>>b;
la=a.size();lb=b.size();
for(int i=0;i<la;i++)
if(i==0)
ai[i]=a[i]-'0';
else
ai[i]=ai[i-1]+a[i]-'0';
for(int i=0;i<lb;i++)
if(i==0)
bi[i]=b[i]-'0';
else
bi[i]=bi[i-1]+b[i]-'0';
}
int check_(int x,int y)
{
if(x%2==y%2)
return 1;
else
return 0;
}
int main()
{
ready();
for(int i=lb-1;i<la;i++)
{
if(i==lb-1)
ans+=check_(bi[lb-1],ai[i]);
else
ans+=check_(bi[lb-1],ai[i]-ai[i-lb]);
}
cout<<ans;
return 0;
}
F题
CodeForces - 33B
Boy Valera likes strings. And even more he likes them, when they are identical. That's why in his spare time Valera plays the following game. He takes any two strings, consisting of lower case Latin letters, and tries to make them identical. According to the game rules, with each move Valera can change one arbitrary character Ai in one of the strings into arbitrary character Bi, but he has to pay for every move a particular sum of money, equal to Wi. He is allowed to make as many moves as he needs. Since Valera is a very economical boy and never wastes his money, he asked you, an experienced programmer, to help him answer the question: what minimum amount of money should Valera have to get identical strings.
Input
The first input line contains two initial non-empty strings s and t, consisting of lower case Latin letters. The length of each string doesn't exceed 105. The following line contains integer n (0 ≤ n ≤ 500) — amount of possible changings. Then follow n lines, each containing characters Ai and Bi (lower case Latin letters) and integer Wi (0 ≤ Wi ≤ 100), saying that it's allowed to change character Ai into character Bi in any of the strings and spend sum of money Wi.
Output
If the answer exists, output the answer to the problem, and the resulting string. Otherwise output -1 in the only line. If the answer is not unique, output any.
Examples
Input
uayd uxxd 3 a x 8 x y 13 d c 3
Output
21 uxyd
Input
a b 3 a b 2 a b 3 b a 5
Output
2 b
Input
abc ab 6 a b 4 a b 7 b a 8 c b 11 c a 3 a c 0
Output
-1
题意：有n种替换字母的方式，将两个只包括小写字母的字符串替换成相同字符串所需要的最少钱是多少，结果字符串是什么。
一开始以为直接逐位比较变成对方就OK了，然后发现可能可以同时变为第三个字母，然后加了个跳板一起变成第三个。然后想想，可以构造一个图，用Floyd来做，两个字母之间有转换也就是有路，费用就是权值，Floyd就能找到两个之间转换到哪个字母比较好，花钱比较少。后面就简单了。
代码：
#include <iostream>
#include <stdio.h>
#include <algorithm>
#include <stdlib.h>
#include <cstdio>
#include <cstdlib>
#include <string>
#include <cstring>
#include <string.h>
#include <math.h>
#include <cmath>
#include <stack>
#include <queue>
#include <list>
#include <vector>
using namespace std;
string a,b;
long long val[30][30];
int len_a,len_b,n;
long long money;
char ans[100005];
void ready()
{
ios::sync_with_stdio(false),cin.tie(0);
cin>>a>>b>>n;
len_a=a.size();
len_b=b.size();
for(int i=1;i<=26;i++)
for(int j=1;j<=26;j++)
val[i][j]=1000000000;
for(int i=1;i<=n;i++)
{
char x,y;
long long v;
cin>>x>>y>>v;
val[x-'a'+1][y-'a'+1]=min(val[x-'a'+1][y-'a'+1],v);
}
for(int i=1;i<=26;i++)
val[i][i]=0;
for(int k=1;k<=26;k++)
for(int i=1;i<=26;i++)
for(int j=1;j<=26;j++)
val[i][j]=min(val[i][j],val[i][k]+val[k][j]);
}
bool work()
{
if(len_a!=len_b)
return false;
for(int i=0;i<len_a;i++)
{
if(a[i]!=b[i])
{
long long cost=1000000000,ci;
int ai=a[i]-'a'+1;
int bi=b[i]-'a'+1;
for(int j=1;j<=26;j++)
{
if(val[ai][j]+val[bi][j]<cost)
{
cost=val[ai][j]+val[bi][j];
ci=j;
}
}
if(cost==1000000000)
return false;
money+=cost;
ans[i]=char('a'+ci-1);
}
else
{
ans[i]=a[i];
}
}
cout<<money<<'\\n';
for(int i=0;i<len_a;i++)
cout<<ans[i];
return true;
}
int main()
{
ready();
if(!work())
cout<<-1;
return 0;
}
G题
Gym - 100962A
Problem A. ABBA Input file: standard input Output file: standard output Time limit: 1 second Memory limit: 256 mebibytes In this problem, we operate with tables of fixed size h × w consisting of real values. Let’s define an addition operation on two tables as their component-wise sum. A multiplication table for two real vectors α = (α1, α2, . . . , αh) and β = (β1, β2 . . . , βw) is the table Tα,β where the element at the intersection of i-th row and j-th column is αi · βj . You start with a table of size h × w consisting of zeroes. In one turn, you are allowed to add a multiplication table for two arbitrary real vectors α of length h and β of length w to the current table. Your task is to make the current table equal to a goal table G in the minimum number of turns. What is the minimum number of turns you have to perform? Input The first line of input contains two integers h and w (1 ≤ h, w ≤ 200). The i-th of the following h lines contain w space-separated integers ai,1, ai,2, . . . , ai,w (−106 ≤ ai,j ≤ 106 ), where ai,j is the value on the intersection of i-th row and j-th column of the goal table G. Output If it’s impossible to obtain the goal table G, print “-1” (without the quotes). Otherwise, output the minimum number of turns you have to perform in order to achieve it. Examples standard input standard output 3 5 1 2 3 4 5 2 4 6 8 10 3 6 9 12 15 1 3 3 2 0 2 0 2 0 2 0 2 2 Note In the first sample, the table T can be obtained using α = ( 1 2 3) , β = ( 1 2 3 4 5) . In the second sample, the table T can be obtained as sum of Tα1,β1 = 1 1 1 1 1 1 1 1 1 for vectors α1 = ( 1 1 1) , β1 = ( 1 1 1) and Tα2,β2 = 1 −1 1 −1 1 −1 1 −1 1 for vectors α2 = ( −1 1 −1 ) , β2 = ( −1 1 −1 )
题意：求矩阵的秩。
身为一个数学与应用数学专业的学生，高等代数上学期也学了，高斯消元法求矩阵的秩也是我们平常的做法。上学期高代学得不错，专业第5，但是这道题看不懂题根本不知道再求秩......
而且以前不敢写高斯消元法的代码就是怕经度问题，现在打出来了，刚刚好可以用进程序设计的课程设计里。nice！
高斯消元法，也就是凑1消去后面的项，直到形成上三角矩阵。
代码：
#include <iostream>
#include <stdio.h>
#include <algorithm>
#include <stdlib.h>
#include <cstdio>
#include <cstdlib>
#include <string>
#include <cstring>
#include <string.h>
#include <math.h>
#include <cmath>
#include <stack>
#include <queue>
#include <list>
#include <vector>
using namespace std;
const long double ep=1e-2;
long double a[205][205];
int r,c;
void ready()
{
ios::sync_with_stdio(false),cin.tie(0);
cin>>r>>c;
for(int i=1;i<=r;i++)
for(int j=1;j<=c;j++)
cin>>a[i][j];
}
void work()
{
int row=1,col=1;
for( ;row<=r && col<=c;row++,col++)
{
int max_r=row;
for(int i=row+1;i<=r;i++)
if(abs(a[i][col])>abs(a[max_r][col]))
max_r=i;
if(max_r!=row)
for(int i=row;i<=c;i++)
swap(a[row][i],a[max_r][i]);
if(abs(a[row][col])<ep)
{
row--;
continue;
}
for(int i=row+1;i<=r;i++)
if(abs(a[i][col])>ep)
{
double t=a[i][col]/a[row][col];
for(int j=row;j<=c;j++)
a[i][j]-=t*a[row][j];
}
}
cout<<row-1;
}
int main()
{
ready();
work();
return 0;
}
好好学英语。下次带字典去了。
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