A.Easy h-index
题目描述
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 , find the h-index of Bobo.
Bobo has published many papers. Given a0, a1, a2 ,..., an which means Bobo has published a i papers with citations exactly i , find the h-index of Bobo.
输入
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 .
The first line of each test case contains an integer n . The second line contains ( n + 1) integers a0 , a1 , ... , an .
输出
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.
样例输入
1 1 2 2 1 2 3 3 0 0 0 0
样例输出
1 2 0 题意：给定被引用次数为0~n的论文分别有几张，找到最大的h，满足被引用次数大于等于h的论文至少有h张思路：在区间[0,n]内二分答案；或直接从n~0遍历找到第一个满足条件的hAC代码：
#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的值可能会爆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;
}
B.Higher h-index
题目描述
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, find the maximum h-index of him.
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, find the maximum h-index of him.
输入
The input consists of several test cases and is terminated by end-of-file.
Each test case contains two integers n and a .
Each test case contains two integers n and a .
输出
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 .
• 1 ≤ n ≤ 109
• 0 ≤ a ≤ n
• The number of test cases does not exceed 104 .
样例输入
3 0 3 1 1000000000 1000000000
样例输出
1 2 1000000000
提示
For the first 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.
题意：给定n个小时，可以用其中x(1<=x<=n)个小时写一篇论文，那么这篇论文的"既定"引用数将会是x*a(a为给定正整数)；此外，已经写好的论文将会被其后写成的论文所引用，也就是说，这篇论文的总引用数将会是"既定"引用数+其后论文篇数；问在所有的写论文方案中(例如一种方案就是用n个小时写n篇论文，每篇论文各花1小时(可以得到这n篇论文的引用数))，h最大为多少(h的含义同上题)(每一种方案都对应着一个h，求这些h中的最大者)
思路：最优方案(即对应h值最大的方案)是平摊n小时写成n篇论文(证明未知)；此时n篇论文的引用数为a,a+1,a+2,...,a+n-1，引用数为a+i时，引用数大于等于它的论文有n-i篇，令a+i=n-i得i=(n-a)/2,所以h=a+(n-a)/2;
AC代码：
#include<cstdio>
int main(){
int n,a;
while(scanf("%d%d",&n,&a)!=EOF){
printf("%d\n",a+(n-a)/2);
}
return 0;
}
C.Just h-index
D.Circular Coloring
E.From Tree to Graph
F.Sorting
题目描述
Bobo has n tuples ( a1 , b1 , c1 ) , ( a2 , b2 , c2 ) , . . . , ( an , bn , cn ). He would like to find the lexicographically smallest permutation p1 , p2 , . . . , pn of 1 ,2 , . . . , n such that for i ∈ {2 , 3 , . . . , n} it holds that
输入
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 .
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 .
输出
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 .
Constraint
• 1 ≤ n ≤ 103
• 1 ≤ ai , bi , ci ≤ 2 × 109
• The sum of n does not exceed 104 .
样例输入
2 1 1 1 1 1 2 2 1 1 2 1 1 1 3 1 3 1 2 2 1 3 1 1
样例输出
2 1 1 2 1 2 3题意：给定n个元组(a1,b1,c1),(a2,b2,c2),...,(an,bn,cn)，将其按(ai+bi)/(ai+bi+ci)的值从小到大排序，输出排序后的n个元组的原序号；思路：编写sort里的cmp函数(形参为元组结构体元素，设为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)再比较大小，这两个结果会爆unsigned long long；可以把因式乘积展开成多项式的和，约去两式中相同的项，得到x.a*y.c+x.b*y.c和y.a*x.c+y.b*x.c，因此只需计算它俩再比较即可，结果不会爆unsigned long long AC代码：
#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;
}
G.String Transformation
题目描述
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 .
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 .
输入
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 .
The first line of each test case contains a string s1 s2 ...sn . The second line contains a string t1 t2 . . . tm .
输出
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.
• 1 ≤ n, m ≤ 104
• s1 , s2 ,..., sn , t1 , t2 , . . . , tm ∈ { a , b , c }
• The sum of n and m does not exceed 250,000.
样例输入
ab ba ac ca a ab
样例输出
Yes No No
提示
For the first sample, Bobo can transform as ab => aababb => babb => ba .
题意：给定字符串S和T(均只由a,b,c三个字母组成)，可对S进行插入或删除操作(插入或删除的子串只能是"aa","bb"或"abab")，问能否通过操作将S变为T
思路：发现经过操作一定可将ab—>ba、ba—>ab(等价于相邻的ab可以互换位置)，那么我们以c为分隔点将S划分为若干小段，对每一个小段，一定可以通过互换相邻a,b的操作使a全在一边，b全在一边，两两约去a，两两约去b，得到“化简”后的小段(可能其中有一个"ab"或有一个'a'或有一个'b'或为空)；就这样分别将每一个小段都“化简”，而原来c的位置不动，得到整个化简后的S；对T进行同样的“化简”操作；若化简后的S，T相同，则Yes，否则No
AC代码：
#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;
}
H.Infinity
I.Longest Increasing Subsequence
J.Vertex Cover
K.2018
题目描述
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.
输入
The input consists of several test cases and is terminated by end-of-file.
Each test case contains four integers a, b, c, d
Each test case contains four integers a, b, c, d
输出
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 .
• 1 ≤ a ≤ b ≤ 109 , 1 ≤ c ≤ d ≤ 109
• The number of tests cases does not exceed 104 .
样例输入
1 2 1 2018 1 2018 1 2018 1 1000000000 1 1000000000
样例输出
3 6051 1485883320325200题意：给定区间[a,b]、[c,d]，问有多少对有序数组(x,y)(x∈[a,b],y∈[c,d])使得x*y是2018的倍数思路：2018=2*1009(分解质因数)，则对x分类讨论:1)仅为2的倍数；2）仅为1009的倍数；3）即为2又为1009的倍数；4）既不为2又不为1009的倍数等价于如下分类讨论：1.若x是偶数：1）若x是1009的倍数，则y可为[c,d]中任意数； 2）若x不是1009的倍数，则y必定为[c,d]中1009的倍数2.若x是奇数：1）若x是1009的倍数，则y必定为[c,d]中2的倍数； 2）若x不是1009的倍数，则y必定为[c,d]中2018的倍数AC代码：
#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;
}
来源：https://www.cnblogs.com/lllxq/p/9057738.html