2018 ccpc 湘潭邀请赛杭电oj

杭电6276：点我
A.Easy h-index
比赛题目：
http://acm.hdu.edu.cn/downloads/2018ccpc_hn.pdf

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 ai papers with citations exactly i, find the h-index of Bobo.

Input
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.

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

Constraint

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

Sample Input
1
1 2
2
1 2 3
3
0 0 0 0

Sample Output
1
2
0

记录一下后缀和

#include<iostream>
#include<cstdio>
using namespace std;

int main(){
    int t;
    while(scanf("%d",&t)!=EOF){
        int sum=0;
        int *a=new int[t+1];
        for(int i=0;i<=t;i++){
            scanf("%d",&a[i]);
        }
        for(int i=t;i>=0;i--){
            sum+=a[i];
            if(sum>=i){
                printf("%d\n",i);
                break;
            }
        }
    }
}

杭电6277：点我
B Higher h-index
Problem Description
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, 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.

Input
The input consists of several test cases and is terminated by end-of-file.

Each test case contains two integers n and a.

Output
For each test case, print an integer which denotes the maximum h-index.

Constraint

1≤n≤109
0≤a≤n
The number of test cases does not exceed 104.

Sample Input
3 0
3 1
1000000000 1000000000

Sample Output
1
2
1000000000

Hint

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 $2$ .

至少有h个论文引用次数不低于h,所以要论文量尽量多，平摊n小时写成n篇论文；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，设h为a+i,令a+i=n-i得i=(n-a)/2,所以h=a+(n-a)/2=(n+a)/2;

#include<iostream>
int main(){
    int a,b;
    while(scanf("%d %d",&a,&b)!=EOF){
        printf("%d\n",(a+b)/2);
    }
}

杭电6281：点我
F.Sorting
Problem Description
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
api−1+bpi−1api−1+bpi−1+cpi−1≤api+bpiapi+bpi+cpi.

Input
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.

Output
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.

Sample Input
2
1 1 1
1 1 2
2
1 1 2
1 1 1
3
1 3 1
2 2 1
3 1 1

Sample Output
2 1
1 2
1 2 3
可以开long double 过

#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
struct node{
    long double fen;
    int index;
};
int cmp(node a,node b){
    if(a.fen==b.fen){
        return a.index<b.index;
    }else{
        return a.fen<b.fen;
    }
}

int main(){
    int n;
    while(scanf("%d",&n)!=EOF){
        node a[1010];
        int d,b,c,i;
        for(i=0;i<n;i++){
            scanf("%d %d %d",&d,&b,&c);
            long double t=d*1.0;
            t+=b*1.0;
            a[i].fen=t/(t+c*1.0);
            a[i].index=i+1;
        } 
        sort(a,a+n,cmp);
        for(int i=0;i<n;i++){
            printf("%d%c",a[i].index,i==n-1?'\n':' ');
        }
    }
}

还可以化成乘积，如x.a+x.b/（x.a+x.b+x.c）与y.a+y.b/（y.a+y.b+y.c）比较，化成乘积（x.a+x.b）（y.a+y.b+y.c）和（y.a+y.b）（x.a+x.b+x.c）比较，实际就是通分，但是会爆掉unsigned long long ,这样的话就把乘积化简，得到x.a*y.c+x.b*y.c和y.a*x.c+y.b*x.c，这样就不会爆unsigned long long 啦

#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;
}

杭电 6282：点我
G.String Transformation

分析：发现大佬们真是啥都会啊，这个题目一看，大佬就知道用异或，主要是消掉aa,bb，就是两个字符串简化后，要一样，但是ab与ba没关系，因为是可以相互转化的，ab中间插入一个abab再简化就是ba了，所以最重要的还是字母c,个数要一样，两个字符串简化后c的相对位置要一样，所以只有记录以c，和以c分段的集合即可，化简后为a,b,ab,ba,只需统计a,b数量即可，所以用到异或。

#include<iostream>
#include<cstdio>
using namespace std;
string count(string s){
    string mod="";
    char sum='a';
    for(int i=0;s[i];i++){
        if(s[i]=='c'){
            mod+=sum,sum='a';
        }
        else{
            sum=sum^s[i];
        }
    }
    mod+=sum;

    return mod;
}
int main(){
    string s1,s2;
    while(cin>>s1>>s2){

        puts(count(s1).compare(count(s2))==0?"Yes":"No");

    }
}

其余的以后再更
这里写图片描述