2018湘潭邀请赛 AFK题解 其他待补...
A.HDU6276:Easy h-index
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1181 Accepted Submission(s): 415
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.
The first line of each test case contains an integer n.
The second line contains (n+1) integers a0,a1,…,an.
## Constraint
* 1≤n≤2⋅105
* 0≤ai≤109
* The sum of n does not exceed 250,000.
#include <stdio.h>
#include <string.h>
#include <string>
#include <map>
#include <queue>
#include <iostream>
#include <algorithm>
#define ll long long
#define exp 1e-8
using namespace std;
const int N = 2e5 +;
const int INF = 2e9+;
const int mod = 1e9+;
ll a[N];
int main() {
int n;
while (~scanf("%d",&n)){
for (int i = ; i <= n; i++) {
scanf("%lld",&a[i]);
}
ll ans = ;
for (int i =n; i >= ; i--){
ans += a[i];
if (ans >= i){
printf("%d\n",i);
break ;
}
}
}
return ;
}
F.HDU6281:Sorting
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1482 Accepted Submission(s): 407
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

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.
DO NOT print trailing spaces.
## Constraint
* 1≤n≤103
* 1≤ai,bi,ci≤2×109
* The sum of n does not exceed 104.
#include <stdio.h>
#include <string.h>
#include <string>
#include <map>
#include <queue>
#include <iostream>
#include <algorithm>
#define ll long long
#define exp 1e-8
using namespace std;
const int N = 1e3 +;
const int INF = 2e9+;
const int mod = 1e9+;
struct node{
int id;
ll a,b,c;
}p[N];
bool cmp(node x,node y){
ll ans1 = (x.a+x.b)*y.c;
ll ans2 = (y.a+y.b)*x.c;
if (ans1 == ans2){
return x.id<y.id;
}
return ans1 < ans2;
}
int main() {
int n;
while (~scanf("%d",&n)){
for (int i = ; i < n; i++) {
p[i].id = i + ;
scanf("%lld%lld%lld",&p[i].a,&p[i].b,&p[i].c);
}
sort(p,p+n,cmp);
for (int i = ; i < n-; i++){
printf("%d ",p[i].id);
}
printf("%d\n",p[n-].id);
}
return ;
}
K.HDU6286:
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 908 Accepted Submission(s): 457
Each test case contains four integers a,b,c,d.
## Constraint
* 1≤a≤b≤109,1≤c≤d≤109
* The number of tests cases does not exceed 104.
#include <stdio.h>
#include <string.h>
#include <string>
#include <map>
#include <queue>
#include <iostream>
#include <algorithm>
#define ll long long
#define ull unsigned ll
#define exp 1e-8
using namespace std;
const int N = 1e3 +;
const int INF = 2e9+;
const int mod = 1e9+;
int main() {
ll a,b,c,d;
while (cin>>a>>b>>c>>d){
ll ans = ;
ll x1 =b/ - (a-)/; //a~b中2018的倍数 × all
ans += x1 * (d-c+);
ll x2 = b/-(a-)/ - x1; //a~b中偶数(除了x1)的个数 × 1009倍数
ans += x2 * (d/ - (c-)/);
ll x3 = b/ - (a-)/ - x1;//a~b中1009的奇数倍 × 偶数
ans += x3 * (d/ - (c-)/);
ll x4 = (b-a+) - (b/-(a-)/) - x3;//a~b中奇数的个数 × 2018倍数
ans += x4 * (d/ - (c-)/);
//printf("%lld %lld %lld %lld \n",x1,x2,x3,x4);
cout << ans << '\n';
}
return ;
}
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