+= 遍歷 我們 find rsa sorting 描述 spec GC

A.Easy h-index

題目描述

The h-index of an author is the largest h where he has at least h papers with citations not less than h.
Bobo has published many papers. Given a0, a1, a2 ,..., an which means Bobo has published a i papers with citations exactly i , ?nd the h-index of Bobo.

輸入

The input consists of several test cases and is terminated by end-of-file.
The first line of each test case contains an integer n . The second line contains ( n + 1) integers a0 , a1 , ... , an .

輸出

For each test case, print an integer which denotes the result.

• 1 ≤ n ≤ 2 · 10⁵
• 0 ≤ ai ≤ 10⁹
• The sum of n does not exceed 250,000.

樣例輸入

1
1 2
2
1 2 3
3
0 0 0 0

樣例輸出

1
2
0

題意：給定被引用次數為0~n的論文分別有幾張，找到最大的h，滿足被引用次數大於等於h的論文至少有h張
思路：在區間[0,n]內二分答案；或直接從n~0遍歷找到第一個滿足條件的h
AC代碼：

#include <iostream>
#include<cstdio>
#include<algorithm>
typedef long long ll;
using namespace std;

ll n;
ll a[200010];

bool ok(ll x){
  ll cnt
 
=0;//cnt的值可能會爆int
  for(ll i=0;i<=n;i++) if(i>=x) cnt+=a[i];
  if(cnt>=x) return true;
  return false;
}

int main()
{
    while(cin>>n){
        for(ll i=0;i<=n;i++){
            cin>>a[i];
        }
        ll l=0,r=n;
        while(l<=r){
            ll mid=(l+r)/2;
            if(ok(mid)) l=mid+1;
            else r=mid-1;
        }
        cout<<r<<endl;
    }
    return 0;
}

View Code

B.Higher h-index

題目描述

The h-index of an author is the largest h where he has at least h papers with citations not less than h .
Bobo has no papers and he is going to publish some subsequently. If he works on a paper for x hours, the paper will get (a·x ) citations from other other persons， where a is a known constant. It’s clear that x should be a positive integer.
There is also a trick – one can cite his own papers published earlier.
Given Bobo has n working hours, ?nd the maximum h-index of him.

輸入

The input consists of several test cases and is terminated by end-of-file.
Each test case contains two integers n and a .

輸出

For each test case, print an integer which denotes the maximum h-index.
• 1 ≤ n ≤ 10⁹
• 0 ≤ a ≤ n
• The number of test cases does not exceed 10⁴ .

樣例輸入

3 0
3 1
1000000000 1000000000

樣例輸出

1
2
1000000000

提示

For the ?rst sample, Bobo can work 3 papers for 1 hour each. With the trick mentioned, he will get papers with citations 2 , 1 , 0. Thus, his h-index is 1.
For the second sample, Bobo can work 2 papers for 1 and 2 hours respectively. He will get papers with citations 1 + 1 , 2 + 0. Thus, his h-index is 2.

題意：給定n個小時，可以用其中x(1<=x<=n)個小時寫一篇論文，那麽這篇論文的"既定"引用數將會是x*a(a為給定正整數)；此外，已經寫好的論文將會被其後寫成的論文所引用，也就是說，這篇論文的總引用數將會是"既定"引用數+其後論文篇數；問在所有的寫論文方案中(例如一種方案就是用n個小時寫n篇論文，每篇論文各花1小時(可以得到這n篇論文的引用數))，h最大為多少(h的含義同上題)(每一種方案都對應著一個h，求這些h中的最大者)

思路：最優方案(即對應h值最大的方案)是平攤n小時寫成n篇論文(證明未知)；此時n篇論文的引用數為a,a+1,a+2,...,a+n-1，引用數為a+i時，引用數大於等於它的論文有n-i篇，令a+i=n-i得i=(n-a)/2,所以h=a+(n-a)/2;

AC代碼：

#include<cstdio>

int main(){
   int n,a;
   while(scanf("%d%d",&n,&a)!=EOF){
    printf("%d\n",a+(n-a)/2);
   }
   return 0;
}

View Code

C.Just h-index

D.Circular Coloring

E.From Tree to Graph

F.Sorting

題目描述

Bobo has n tuples ( a1 , b1 , c1 ) , ( a2 , b2 , c2 ) , . . . , ( an , bn , cn ). He would like to ?nd the lexicographically smallest permutation p1 , p2 , . . . , pn of 1 ,2 , . . . , n such that for i ∈ {2 , 3 , . . . , n} it holds that

輸入

The input consists of several test cases and is terminated by end-of-file.
The first line of each test case contains an integer n . The i-th of the following n lines contains 3 integers ai,bi and ci .

輸出

For each test case, print n integers p1, p2,..., pn seperated by spaces. DO NOT print trailing spaces.
Constraint
• 1 ≤ n ≤ 10³
• 1 ≤ ai , bi , ci ≤ 2 × 10⁹
• The sum of n does not exceed 10⁴ .

樣例輸入

2
1 1 1
1 1 2
2
1 1 2
1 1 1
3
1 3 1
2 2 1
3 1 1

樣例輸出

2 1
1 2
1 2 3

題意：給定n個元組(a1,b1,c1),(a2,b2,c2),...,(an,bn,cn)，將其按(ai+bi)/(ai+bi+ci)的值從小到大排序，輸出排序後的n個元組的原序號；
思路：編寫sort裏的cmp函數(形參為元組結構體元素，設為Tuple x,Tuple y)，若直接算出(x.a+x.b)*(y.a+y.b+y.c)和(y.a+y.b)*(x.a+x.b+x.c)再比較大小，這兩個結果會爆unsigned long long；
可以把因式乘積展開成多項式的和，約去兩式中相同的項，得到x.a*y.c+x.b*y.c和y.a*x.c+y.b*x.c，因此只需計算它倆再比較即可，結果不會爆unsigned long long 
AC代碼：

#pragma GCC optimize(2)
#include<iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <set>
using namespace std;

struct Tuple{
  long long a,b,c;
  int ind;
}tup[100010];

bool cmp(Tuple x,Tuple y){
  unsigned long long l=x.a*y.c+x.b*y.c;
  unsigned long long r=y.a*x.c+y.b*x.c;
  if(l==r) return x.ind<y.ind;
  return l<r;
}

int main(){
  int n;
  while(scanf("%d",&n)!=EOF){
    for(int i=1;i<=n;i++){
        cin>>tup[i].a>>tup[i].b>>tup[i].c;
        tup[i].ind=i;
    }
    sort(tup+1,tup+1+n,cmp);
    for(int i=1;i<=n;i++){
        if(i!=1) printf(" ");
        cout<<tup[i].ind;
    }
    printf("\n");
  }
  return 0;
}

View Code

G.String Transformation

題目描述

Bobo has a string S = s1 s2...sn consists of letter a , b and c . He can transform the string by inserting or deleting substrings aa , bb and abab .
Formally, A = u ? w ? v (“ ? ” denotes string concatenation) can be transformed into A 0 = u ? v and vice versa where u , v are (possibly empty) strings and w ∈ { aa , bb , abab } .
Given the target string T = t1 t2 . . . tm , determine if Bobo can transform the string S into T .

輸入

The input consists of several test cases and is terminated by end-of-file.
The first line of each test case contains a string s1 s2 ...sn . The second line contains a string t1 t2 . . . tm .

輸出

For each test case, print Yes if Bobo can. Print No otherwise.
• 1 ≤ n, m ≤ 10⁴
• s1 , s2 ,..., sn , t1 , t2 , . . . , tm ∈ { a , b , c }
• The sum of n and m does not exceed 250,000.

樣例輸入

ab
ba
ac
ca
a
ab

樣例輸出

Yes
No
No

提示

For the first sample, Bobo can transform as ab => aababb => babb => ba .

題意：給定字符串S和T(均只由a,b,c三個字母組成)，可對S進行插入或刪除操作(插入或刪除的子串只能是"aa","bb"或"abab")，問能否通過操作將S變為T

思路：發現經過操作一定可將ab—>ba、ba—>ab(等價於相鄰的ab可以互換位置)，那麽我們以c為分隔點將S劃分為若幹小段，對每一個小段，一定可以通過互換相鄰a,b的操作使a全在一邊，b全在一邊，兩兩約去a，兩兩約去b，得到“化簡”後的小段(可能其中有一個"ab"或有一個‘a‘或有一個‘b‘或為空)；就這樣分別將每一個小段都“化簡”，而原來c的位置不動，得到整個化簡後的S；對T進行同樣的“化簡”操作；若化簡後的S，T相同，則Yes，否則No

AC代碼：

#include <iostream>
#include<cstdio>
#include<cstring>
using namespace std;
 
char s1[1000010];
char s2[1000010];
char tmp1[1000010];
char tmp2[1000010];
 
int main()
{
    while(scanf("%s%s",s1,s2)!=EOF){
        int len=strlen(s1);
        int cnt_1=0,cnt_2=0;
        int cnt_a=0,cnt_b=0;
        for(int i=0;i<len;i++){
            if(s1[i]==‘c‘) {
                if(cnt_a%2==0&&cnt_b%2==0) ;
                else if(cnt_a%2==1&&cnt_b%2==0) tmp1[++cnt_1]=‘a‘;
                else if(cnt_a%2==0&&cnt_b%2==1) tmp1[++cnt_1]=‘b‘;
                else {tmp1[++cnt_1]=‘a‘; tmp1[++cnt_1]=‘b‘;}
                tmp1[++cnt_1]=‘c‘;
                cnt_a=0; cnt_b=0;
            }
            else if(s1[i]==‘a‘) cnt_a++;
            else cnt_b++;
        }
        if(cnt_a%2==0&&cnt_b%2==0) ;
        else if(cnt_a%2==1&&cnt_b%2==0) tmp1[++cnt_1]=‘a‘;
        else if(cnt_a%2==0&&cnt_b%2==1) tmp1[++cnt_1]=‘b‘;
        else {tmp1[++cnt_1]=‘a‘; tmp1[++cnt_1]=‘b‘;}
        cnt_a=0; cnt_b=0;
        len=strlen(s2);
        for(int i=0;i<len;i++){
            if(s2[i]==‘c‘) {
                if(cnt_a%2==0&&cnt_b%2==0) ;
                else if(cnt_a%2==1&&cnt_b%2==0) tmp2[++cnt_2]=‘a‘;
                else if(cnt_a%2==0&&cnt_b%2==1) tmp2[++cnt_2]=‘b‘;
                else {tmp2[++cnt_2]=‘a‘; tmp2[++cnt_2]=‘b‘;}
                tmp2[++cnt_2]=‘c‘;
                cnt_a=0; cnt_b=0;
            }
            else if(s2[i]==‘a‘) cnt_a++;
            else cnt_b++;
        }
        if(cnt_a%2==0&&cnt_b%2==0) ;
        else if(cnt_a%2==1&&cnt_b%2==0) tmp2[++cnt_2]=‘a‘;
        else if(cnt_a%2==0&&cnt_b%2==1) tmp2[++cnt_2]=‘b‘;
        else {tmp2[++cnt_2]=‘a‘; tmp2[++cnt_2]=‘b‘;}
        cnt_a=0; cnt_b=0;
        if(cnt_1!=cnt_2) printf("No\n");
        else{
            int flag=0;
            for(int i=1;i<=cnt_1;i++){
                if(tmp1[i]!=tmp2[i]) {flag=1;break;}
            }
            if(flag) printf("No\n");
            else printf("Yes\n");
        }
    }
    return 0;
}

View Code

H.Infinity

I.Longest Increasing Subsequence

J.Vertex Cover

K.2018

題目描述

Given a, b, c, d , find out the number of pairs of integers ( x, y ) where a ≤ x ≤ b, c ≤ y ≤ d and x · y is a multiple of 2018.

輸入

The input consists of several test cases and is terminated by end-of-file.
Each test case contains four integers a, b, c, d

輸出

For each test case, print an integer which denotes the result.
• 1 ≤ a ≤ b ≤ 10⁹ , 1 ≤ c ≤ d ≤ 10⁹
• The number of tests cases does not exceed 10⁴ .

樣例輸入

1 2 1 2018
1 2018 1 2018
1 1000000000 1 1000000000

樣例輸出

3
6051
1485883320325200

題意：給定區間[a,b]、[c,d]，問有多少對有序數組(x,y)(x∈[a,b],y∈[c,d])使得x*y是2018的倍數
思路：2018=2*1009(分解質因數)，則對x分類討論:1)僅為2的倍數；2）僅為1009的倍數；3）即為2又為1009的倍數；4）既不為2又不為1009的倍數
等價於如下分類討論：
1.若x是偶數：1）若x是1009的倍數，則y可為[c,d]中任意數； 2）若x不是1009的倍數，則y必定為[c,d]中1009的倍數
2.若x是奇數：1）若x是1009的倍數，則y必定為[c,d]中2的倍數； 2）若x不是1009的倍數，則y必定為[c,d]中2018的倍數
AC代碼：

#include<cstdio>
#include<iostream>
typedef unsigned long long ll;
using namespace std;

int main(){
   ll a,b,c,d;
   while(cin>>a>>b>>c>>d){
     ll num1_all_1009=b/1009-(a-1)/1009;
     ll num1_even=b/2-(a-1)/2;
     ll num1_1009_in_even=b/2018-(a-1)/2018;
     ll num1_rest_in_even=num1_even-num1_1009_in_even;
     ll num1_odd=(b-a+1)-num1_even;
     ll num1_1009_in_odd=num1_all_1009-num1_1009_in_even;
     ll num1_rest_in_odd=num1_odd-num1_1009_in_odd;
     ll ans=0;
     ans+=num1_1009_in_even*(d-c+1);
     ll num2_all_1009=d/1009-(c-1)/1009; ans+=num1_rest_in_even*num2_all_1009;
     ll num2_even=d/2-(c-1)/2; ans+=num1_1009_in_odd*num2_even;
     ll num2_all_2018=d/2018-(c-1)/2018; ans+=num1_rest_in_odd*num2_all_2018;
     cout<<ans<<endl;
   }
   return 0;
}

View Code

[題解]2018湘潭邀請賽