HDU 6438
1001 Buy and Resell
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2282 Accepted Submission(s): 359
Problem Description
The Power Cube is used as a stash of Exotic Power. There are n cities numbered 1,2,…,n where allowed to trade it. The trading price of the Power Cube in the i-th city is ai dollars per cube. Noswal is a foxy businessman and wants to quietly make a fortune by buying and reselling Power Cubes. To avoid being discovered by the police, Noswal will go to the i-th city and choose exactly one of the following three options on the i-th day:
1. spend ai dollars to buy a Power Cube
2. resell a Power Cube and get ai dollars if he has at least one Power Cube
3. do nothing
Obviously, Noswal can own more than one Power Cubes at the same time. After going to the n cities, he will go back home and stay away from the cops. He wants to know the maximum profit he can earn. In the meanwhile, to lower the risks, he wants to minimize the times of trading (include buy and sell) to get the maximum profit. Noswal is a foxy and successful businessman so you can assume that he has infinity money at the beginning.
Input
There are multiple test cases. The first line of input contains a positive integer T (T≤250), indicating the number of test cases. For each test case:
The first line has an integer n. (1≤n≤105)
The second line has n integers a1,a2,…,an where ai means the trading price (buy or sell) of the Power Cube in the i-th city. (1≤ai≤109)
It is guaranteed that the sum of all n is no more than 5×105.
Output
For each case, print one line with two integers —— the maximum profit and the minimum times of trading to get the maximum profit.
Sample Input
3 4 1 2 10 9 5 9 5 9 10 5 2 2 1
Sample Output
16 4 5 2 0 0
Hint
In the first case, he will buy in 1, 2 and resell in 3, 4. profit = - 1 - 2 + 10 + 9 = 16 In the second case, he will buy in 2 and resell in 4. profit = - 5 + 10 = 5 In the third case, he will do nothing and earn nothing. profit = 0
貪心，每次保留買，和賣兩種狀态，如果一種買了，說明後面有一定有可以賣的，從前往後掃每次取前面最小的，如果遇見能從賣了的裡面取，總次數不用+2，如果是從買了的裡面取，總交易次數+2；
用，0，1分别代表買和賣。
#include<iostream>
#include<algorithm>
#include<cstring>
#include<string>
#include<vector>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<queue>
#include<map>
#include<set>
#include<stack>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> P;
#define bug printf("*********\n");
#define debug(x) cout<<"["<<x<<"]" <<endl;
#define mid (l+r)/2
#define chl 2*k+1
#define chr 2*k+2
#define lson l,mid,chl
#define rson mid,r,chr
#define pb push_back
#define mem(a,b) memset(a,b,sizeof(a));
const long long mod=1e9+7;
const int maxn=5e5+5;
const int INF=0x7fffffff;
const int inf=0x3f3f3f3f;
const double eps=1e-8;
int a[maxn];
int n;
priority_queue<P>q;
ll sum=0,ans=0;
int main() {
int t;
cin>>t;
while(t--) {
sum=0;
ans=0;
scanf("%d",&n);
while(q.size())q.pop();
for(int i=0; i<n; i++) {
scanf("%d",&a[i]);
q.push(P(-a[i],0));
q.push(P(-a[i],1));
int temp=a[i]+q.top().first;
if(q.top().second==0) {
ans+=2;
}
q.pop();
sum+=temp;
}
printf("%lld %lld\n",sum,ans);
}
return 0;
}
HDU 6441
1004 Find Integer
更具費馬大定理，n>2 和等于 0誤解，n=1，直接輸出 1 a+1;
n=2 的時候，就是一個勾股定理，a^2=c*c-b*b = (c-b)*(c+b) 如果A為偶數 c-b=2 c+b=a*a/2 ,如果a為奇數 c-b=1 c+b=a*a
解個方程就行了。
#include<iostream>
#include<algorithm>
#include<cstring>
#include<string>
#include<vector>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<queue>
#include<map>
#include<set>
#include<stack>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> P;
#define bug printf("*********\n");
#define debug(x) cout<<"["<<x<<"]" <<endl;
#define mid (l+r)/2
#define chl 2*k+1
#define chr 2*k+2
#define lson l,mid,chl
#define rson mid,r,chr
#define pb push_back
#define mem(a,b) memset(a,b,sizeof(a));
const long long mod=1e9+7;
const int maxn=5e5+5;
const int INF=0x7fffffff;
const int inf=0x3f3f3f3f;
const double eps=1e-8;
int main() {
int t;
cin>>t;
while(t--) {
ll n,k;
scanf("%lld%lld",&k,&n);
if(k==0||k>2)puts("-1 -1");
else if(k==1) {
if(n==1e9)puts("-1 -1");
else printf("%lld %lld\n",1,n+1);
} else {
ll temp=n;
if(n==1||n==2)puts("-1 -1");
else if(n&1) {
ll c=(temp*temp+1)/2;
ll b=c-1;
if(c>1e9)puts("-1 -1");
else {
printf("%lld %lld\n",b,c);
}
} else {
ll c=n*n/4+1,b=c-2;
if(c>1e9)puts("-1 -1");
else printf("%lld %lld\n",b,c);
}
}
}
return 0;
}
HDU 6446
1009 Tree and Permutation
直接算一條邊左右兩邊點的個數，全排列種，每條邊經過的次數等于 ，$2*C,(n-m)*(m)*(n-1)！$次，一個DFS求出他根節點的數量為m，前一條邊 權值為C,求一下所有數的和，預處理階乘。
#include<iostream>
#include<algorithm>
#include<cstring>
#include<string>
#include<vector>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<queue>
#include<map>
#include<set>
#include<stack>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> P;
#define bug printf("*********\n");
#define debug(x) cout<<"["<<x<<"]" <<endl;
#define mid (l+r)/2
#define chl 2*k+1
#define chr 2*k+2
#define lson l,mid,chl
#define rson mid,r,chr
#define pb push_back
#define mem(a,b) memset(a,b,sizeof(a));
const long long mod=1e9+7;
const int maxn=5e5+5;
const int INF=0x7fffffff;
const int inf=0x3f3f3f3f;
const double eps=1e-8;
int n;
struct edge {
int to,cost,next;
} eg[maxn];
int head[maxn],tot;
void init() {
mem(head,-1);
tot=0;
}
void add(int u,int v,int c) {
eg[tot].to=v;
eg[tot].cost=c;
eg[tot].next=head[u];
head[u]=tot++;
}
ll k[maxn];
ll c[maxn],num[maxn];
int dfs(int r,int p,int v) {
c[r]=v%mod;
int ans=1;
for(int i=head[r]; i!=-1; i=eg[i].next) {
if(eg[i].to!=p) {
ans+=dfs(eg[i].to,r,eg[i].cost);
}
}
num[r]=ans%mod;
return ans;
}
int main() {
k[1]=1;
for(int i=2; i<=1e5; i++) {
k[i]=k[i-1]*i%mod;
}
while(~scanf("%d",&n)) {
init();
for(int i=1; i<n; i++) {
int u,v,c;
scanf("%d%d%d",&u,&v,&c);
add(u,v,c);
add(v,u,c);
}
dfs(1,-1,0);
ll sum=0;
for(int i=2; i<=n; i++) {
sum+=2*c[i]*num[i]%mod*(n-num[i])%mod*k[n-1]%mod;
sum%=mod;
}
cout<<sum<<endl;
}
return 0;
}
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