题目链接:http://bestcoder.hdu.edu.cn/contests/contest_show.php?cid=863
A.
#include <bits/stdc++.h>
using namespace std;
const int mx = 2e5 + 10;
const int mod = 1e9+7;
typedef long long ll;
int main(){
int t;
scanf("%d",&t);
while(t--){
int n;
scanf("%d",&n);
printf("%d\n",n^1);
}
return 0;
}B.
跑个最短路，然后剪剪枝就OK了。
#include <bits/stdc++.h>
#define x first
#define y second
using namespace std;
const int mx = 1e3 + 10;
const int mod = 998244353;
typedef long long ll;
typedef pair <int,int> pa;
vector <int> r;
vector <pa> g[mx];
ll d[mx][mx],dis[mx];
int c[mx][mx];
int n,m;
bool vis[mx];
void spfa(int beg){
queue <int> q;
for(int i=1;i<=n;i++)
dis[i] = 1e16;
dis[beg] = 0;
q.push(beg);
while(!q.empty()){
int now = q.front();
q.pop();
vis[now] = 0;
for(pa po:g[now]){
int v = po.x,w = po.y;
if(dis[v]>dis[now]+po.y){
dis[v] = dis[now] + po.y;
if(!vis[v]){
vis[v] = 1;
q.push(v);
}
}
}
}
}
int main(){
int t;
scanf("%d",&t);
while(t--){
scanf("%d%d",&n,&m);
for(int i=1;i<=n;i++){
for(int j=1;j<=n;j++){
c[i][j] = 0,d[i][j] = 1e16;
if(i==j) d[i][j] = 0;
}
g[i].clear();
}
r.clear();
for(int i=1;i<=m;i++){
int u,v,w;
scanf("%d%d%d",&u,&v,&w);
d[u][v] = min(1ll*w,d[u][v]);
d[v][u] = min(1ll*w,d[v][u]);
g[v].push_back(pa(u,w));
g[u].push_back(pa(v,w));
}
for(int k=1;k<=n;k++){
spfa(k);
r.clear();
for(int i=1;i<=n;i++){
if(dis[i]==d[k][i])
r.push_back(i);
if(d[k][i]!=d[i][k]) cout << 1/0 << endl;
}
for(int i:r){
for(int j:r){
ll tmp = d[i][k] + d[k][j];
if(tmp<d[i][j]){
d[i][j] = tmp;
c[i][j] = k;
}
}
}
}
ll ans = 0;
for(int i=1;i<=n;i++){
for(int j=1;j<=n;j++){
ans += c[i][j];
//cout << dis[i][j] << endl;
}
}
printf("%I64d\n",ans%mod);
}
return 0;
}C.
利用莫比乌斯函数的性质和u(lcm(i,j)） = u(i)*u(j)*u(gcd(i,j))
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef complex<double> comp;
const double pi = acos(-1);
const int mx = 1e6+10;
const int mod = 1e9 + 7;
int mu[mx],pri[mx/10],pre[mx];
bool vis[mx];
vector <int> g[mx];
void init(){
int top = 0;
mu[1] = 1;
for(int i=2;i<mx;i++){
if(!vis[i]){
pri[top++] = i;
mu[i] = -1;
}
for(int j=0;j<top&&pri[j]*i<mx;j++){
int val = i*pri[j];
vis[val] = 1;
if(i%pri[j]==0){
mu[val] = 0;
break;
}
mu[val] = -mu[i];
}
}
for(int i=1;i<mx;i++){
g[i].push_back(0);
int sum = 0;
for(int j=i;j<mx;j+=i){
sum += mu[j];
g[i].push_back(sum);
pre[j] += mu[i]*mu[j/i];
}
}
}
int main()
{
init();
int t;
scanf("%d",&t);
while(t--){
int n,m;
scanf("%d%d",&n,&m);
int mi = min(n,m);
ll ans = 0;
for(int i=1;i<=mi;i++){
ans += 1ll*pre[i]*g[i][n/i]*g[i][m/i];
}
printf("%lld\n",ans);
}
return 0;
}
