以下所有AC题解程序来自“仙客传奇”团队。
AC的C++语言程序：
#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
#include<vector>
using namespace std;
const int maxn=1e5+10;
priority_queue< int,vector<int>,greater<int> > buy,sell;
typedef long long ll;
//priority_queue 默认从大到小 修改后从小到大
ll val[maxn];
int main()
{
int n,t;
scanf("%d",&t);
while(t--){
scanf("%d",&n);
for(int i=0;i<n;i++)
scanf("%lld",&val[i]);
ll ans=0,cnt=0;//tiaixao????
for(int i=0;i<n;i++){
//cout<<ans<<endl;
if(buy.size()!=0&&sell.size()!=0&&buy.top()<sell.top()&&sell.top()<val[i]){
ans+=val[i];cnt++;
sell.push(val[i]);
buy.pop();
}
else if(sell.size()!=0&&sell.top()<val[i]){//交换
buy.push(sell.top());
sell.pop();
sell.push(val[i]);
ans+=val[i]-2*buy.top();
}
else if(buy.size()!=0&&buy.top()<val[i]) {
ans+=val[i];cnt++;//cout<<i<<endl;
sell.push(val[i]);
buy.pop();
}
else buy.push(val[i]),ans-=val[i];
}
while(buy.size()!=0){
ans+=buy.top();
buy.pop();
}
printf("%lld %lld\n",ans,cnt*2);
while(!buy.empty()) buy.pop();
while(!sell.empty()) sell.pop();
}
return 0;
}
/*
4
8
1 2 3 4 5 6 7 8
4
1 2 10 9
5
9 5 9 10 5
2
2 1*/
题解链接：
HDU 6439 Congruence equation（莫比乌斯反演）
HDU 6439 2018CCPC网络赛 Congruence equationI（杜教筛 + 莫比乌斯反演 + 伯努利数）
题解链接：
2018中国大学生程序设计竞赛 - 网络选拔赛（hdu6440 Dream）
AC的C++语言程序：
#include<iostream>
#include<cmath>
using namespace std;
const int maxn = 4e4;
int main()
{
// ios::sync_with_stdio(false);
// cin.tie(NULL);cout.tie(NULL);
int T,n,a;
scanf("%d",&T);
while(T--)
{
scanf("%d%d",&n,&a);
if(n == 0 || n > 2)
printf("-1 -1\n");
else if(n == 1)
printf("1 %d\n",a + 1);
else if(n == 2)
{
if(a % 2 == 0)
printf("%d %d\n",a * a / 4 - 1, a * a / 4 + 1);
else
printf("%d %d\n",(a - 1) * (a / 2 + 1),(a - 1) * (a / 2 + 1) + 1);
}
}
return 0;
}
题解链接：
HDU 6442 GuGu Convolution（快速幂）
hdu 6442 - 二项式定理
题解链接：
hdu6444（最大子段和+gcd）
hdu 6444 - 最大子段和(单调队列)
题解链接：
HDU 6445(竞赛图 + 网络流)
HDU 6445 2018CCPC网络赛1008 Search for Answer（费用流 + 构图）
AC的C++语言程序：
#include<iostream>
#include<cstring>
#include<vector>
using namespace std;
typedef long long ll;
const int maxn = 1e5;
const int mod = 1e9 + 7;
ll fac[maxn + 10],n,sum;
struct edge
{
int u,v;
ll w;
edge(){}
edge(int _u,int _v,ll _w):u(_u),v(_v),w(_w){}
}e[maxn + 10];
vector<edge> son[maxn + 10];
int cnt[maxn + 10],vis[maxn + 10];
ll fa[maxn + 10];
void getfac()
{
fac[0] = fac[1] = 1;
for(int i = 2;i <= maxn;++i)
fac[i] = (fac[i - 1] * i) % mod;
}
int dfs(int no)
{
int cnt = 1;
vis[no] = 1;
for(int i = 0;i < son[no].size();++i)
if(!vis[son[no][i].v])
{
int temp = dfs(son[no][i].v);
cnt += temp;
sum = (sum + ((son[no][i].w * fac[n - 1] * 2ll) % mod * ((n - temp) * temp) % mod)) % mod;
}
return cnt;
}
int main()
{
ios::sync_with_stdio(false);
cin.tie(NULL);cout.tie(NULL);
getfac();
while(cin>>n)
{
// memset(cnt,0,sizeof(cnt));
memset(vis,0,sizeof(vis));
for(int i = 1;i <= n;++i) son[i].clear();
for(int i = 1;i < n;++i)
{
int u,v,w;
cin>>u>>v>>w;
son[u].push_back(edge(u,v,w));
son[v].push_back(edge(v,u,w));
}
sum = 0;
dfs(1);
cout<<sum<<endl;
}
return 0;
}
AC的C++语言程序：
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#define INF 2147483647
#define MEM(x, b) memset(x, b ,sizeof(x))
#define ll long long
using namespace std;
const int MOD = 1e9 + 7;
const int MAXN = 1e5 + 2;
int n;
vector<pair<int, int> > edge[MAXN];
int num[MAXN];
ll fac[MAXN];
ll ans;
void dfs(int crt, int fa){
for (int i = 0; i < edge[crt].size(); i++) {
int u = edge[crt][i].first;
if (u == fa) continue;
dfs(u, crt);
num[crt] += num[u];
ans =(ans + 1ll*(num[u]) %MOD * (n - num[u]) % MOD* fac[n - 1] % MOD * 2ll % MOD * edge[crt][i].second % MOD) % MOD;
}
}
int main() {
fac[0] = 1;
for (int i = 1; i < MAXN; i++) {
fac[i] = (fac[i - 1] * (long long)i % MOD) % MOD;
}
while (~scanf("%d", &n)){
ans = 0;
for (int i = 1; i <= n; i++) edge[i].clear();
//MEM(num, 1);
for (int i = 1; i<= n; i++) num[i] = 1;
for (int i = 1; i < n; i++){
int u , v, w;
scanf("%d%d%d",&u,&v,&w);
edge[u].push_back(make_pair(v, w));
edge[v].push_back(make_pair(u, w));
}
//cout << "ds" << endl;
dfs(1, 0);
printf("%lld\n", ans);
}
return 0;
}
AC的C++语言程序：
#include <cstdio>
#include <vector>
#include <cstring>
#include <algorithm>
using namespace std;
const int MAXN = 1e5 + 2;
struct node{
int x, y, w;
bool operator < (node A) {
if (x == A.x) {
return y>A.y;
}else return x < A.x;
}
};
node p[MAXN];
int tr[MAXN], a[MAXN];
int n ,tot;
int lowbit(int x){
return (x & (-x));
}
int query(int l) {
int maxn = 0;
for (int i = l; i >=1; i-=lowbit(i)){
maxn = max(maxn, tr[i]);
}
return maxn;
}
void add(int pos , int val) {
tr[pos]= val;
for (int i = pos; i <= MAXN; i+= lowbit(i) ){
tr[i] = max(tr[i], val);
}
}
int main() {
int t;
scanf("%d", &t);
while (t--) {
memset(tr, 0 ,sizeof(tr));
scanf("%d", &n);
for (int i = 0; i < n; i++){
scanf("%d%d%d", &p[i].x, &p[i].y, &p[i].w);
a[i] = p[i].y;
}
sort(a , a + n);
tot = unique(a, a + n) - a;
for (int i = 0; i < n; i++){
p[i].y = lower_bound(a, a + n, p[i].y) - a + 1;
}
sort(p, p + n);
for (int i = 0; i < n; i++){
int val = query(p[i].y - 1) + p[i].w;
add(p[i].y, val);
}
printf("%d\n", query(MAXN - 1)) ;
}
}
题解链接：
BEIL 2018 ACM-ICPC 中国大学生程序设计竞赛线上赛 Goldbach
ACDIJ 2018中国大学生程序设计竞赛网络选拔赛