A
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
string st;
int main(){
//freopen("in.txt","r",stdin);
cin>>st;
int cnt=0;
for(int i=0;i<(int)st.size();i++)
for(int j=i+1;j<(int)st.size();j++)
for(int k=j+1;k<(int)st.size();k++)
if(st[i]=='Q'&&st[j]=='A'&&st[k]=='Q')
cnt++;
cout<<cnt;
return 0;
}
B
猜结论。。。
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
ll n,m,k;
inline ll power(ll a,ll n,ll p){
ll ret=1;ll now=a;
while(n!=0){
if(n&1)
ret=ret*now%p;
now=now*now%p;
n>>=1;
}
return ret;
}
int main(){
//freopen("in.txt","r",stdin);
cin>>n>>m>>k;
if(((n+m)&1)&&k==-1)
puts("0");
else {
ll res = power(2,n-1,1000000007);
res = power(res,m-1,1000000007);
cout<<res;
}
return 0;
}
C
提供了一个假算法，惨遭hack。正解构造。
如果整个数列的最大公约数是最小的那个数字，那么可以构造出一个数列，只需要让每个数字中间夹着一个a[0] 就可以让区间长度大于等于二的区间的最大公约数为a[0]。
#include <stdio.h>
int n,a[1005];
int main(){
scanf("%d",&n);
for(int i=0;i<n;i++) scanf("%d",a+i);
bool flg=1;
for(int i=0;i<n;i++) if(a[i]%a[0]) flg=0;
if(!flg) printf("-1");
else {printf("%d
",2*n);for(int i=0;i<n;i++) printf("%d %d ",a[i],a[0]);}
return 0;
}
D
预处理出每个子树，每个点到该子树的根的距离，排个序。复杂度$O(n log n log n)$
然后对于每个询问，每次向上爬就行了。复杂度$O(m log n log n)$。
#include <bits/stdc++.h>
#define pb push_back
using namespace std;
typedef long long ll;
const int maxn = 1e6 + 5;
vector<int> dis[maxn];
vector<ll> sum[maxn];
int n, m, A;
ll H;
int L[maxn];
int cnt;
template<typename T> inline void read(T &x) {
x = 0; T f = 1; char ch; do {ch = getchar(); if (ch == '-')f = -1;} while (ch < '0' || ch > '9'); do x = x * 10 + ch - '0', ch = getchar(); while (ch <= '9' && ch >= '0'); x *= f;
}
template<typename A, typename B> inline void read(A&x, B&y) {read(x); read(y);}
template<typename A, typename B, typename C> inline void read(A&x, B&y, C&z) {read(x); read(y); read(z);}
template<typename A, typename B, typename C, typename D> inline void read(A&x, B&y, C&z, D&w) {read(x); read(y); read(z); read(w);}
void dfs(int id) {
dis[id].pb(0);
if (id * 2 > n) {
return;
} else if (id * 2 + 1 > n) {
int l = id * 2;
dfs(l);
for (auto d : dis[l]) {
dis[id].pb(d + L[l]);
}
} else {
int l = id * 2, r = id * 2 + 1;
dfs(l);
for (auto d : dis[l]) {
dis[id].pb(d + L[l]);
}
dfs(r);
for (auto d : dis[r]) {
dis[id].pb(d + L[r]);
}
}
}
ll get(int A, ll H) {
int id = upper_bound(dis[A].begin(), dis[A].end(), H) - dis[A].begin();
if (!id) return 0;
return H * id - sum[A][id - 1];
}
int main() {
//freopen("in.txt","r",stdin);
read(n, m);
for (int i = 2; i <= n; i++)
read(L[i]);
dfs(1);
for (int i = 1; i <= n; i++)
sort(dis[i].begin(), dis[i].end());
for (int i = 1; i <= n; i++) {
ll tmp = 0;
for (int j = 0; j < (int)dis[i].size(); j++) {
tmp += dis[i][j];
sum[i].pb(tmp);
}
}
ll xres;
while (m--) {
read(A, H);
ll res = 0;
res += xres = get(A, H);
//cout<<"#"<<xres<<endl;
int lst = A;
int now = A / 2;
while (now) {
H -= L[lst];
if (H < 0) break;
res += H;
if (lst & 1) {
if (H >= L[now * 2]) {
res += xres = get(now * 2, H - L[now * 2]);
//cout<<"#"<<xres<<endl;
}
} else {
if (H >= L[now * 2 + 1]) {
res += xres = get(now * 2 + 1, H - L[now * 2 + 1]);
//cout<<"#"<<xres<<endl;
}
}
lst = now;
now = now / 2;
}
printf("%I64d
", res);
}
return 0;
}
E
首先，强连通分量中的每条边是都可以走"完"的。
对于边权w，可以二分一个最大的p，满足$w-1-2-...-p>=0$，此时的贡献为$w(p+1)-frac{p(p+1)(p+2)}{6}$.
缩点之后做一次dag上的dp。
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 1e6+5;
const int maxm = 1e6+5;
template<typename T> inline void read(T &x){
x=0;T f=1;char ch;do{ch=getchar();if(ch=='-')f=-1;}while(ch<'0'||ch>'9');do x=x*10+ch-'0',ch=getchar();while(ch<='9'&&ch>='0');x*=f;
}
template<typename A,typename B> inline void read(A&x,B&y){read(x);read(y);}
template<typename A,typename B,typename C> inline void read(A&x,B&y,C&z){read(x);read(y);read(z);}
template<typename A,typename B,typename C,typename D> inline void read(A&x,B&y,C&z,D&w){read(x);read(y);read(z);read(w);}
struct node{
int v;ll w;
node(){}
node(int _v,ll _w):v(_v),w(_w){}
};
vector<node> e[maxn];
int Low[maxn],DFN[maxn],Stack[maxn],Belong[maxn];
int num[maxn];
int Index,top,n,m,x,y;
ll z;
int scc;
bool Instack[maxn];
ll score[maxn];
vector<int> scce[maxn];
ll dp[maxn];
void Tarjan(int u){
int v;
Low[u]=DFN[u]=++Index;
Stack[top++]=u;
Instack[u]=true;
for(auto enxt:e[u]){
v=enxt.v;
if(!DFN[v]){
Tarjan(v);
Low[u]=min(Low[u],Low[v]);
} else if(Instack[v] && Low[u]>DFN[v])
Low[u]=DFN[v];
}
if(Low[u]==DFN[u]){
scc++;
do{
v=Stack[--top];
Instack[v]=false;
Belong[v]=scc;
num[scc]++;
}while(v!=u);
}
}
int indeg[maxn],outdeg[maxn];
ll calc(ll w){
ll l=0,r=1e6+5,rt;
while(l<=r){
ll mid=(l+r)/2;
if(mid*(mid+1)/2<=w)
l=(rt=mid)+1;
else r=mid-1;
}
ll ret=w*(rt+1ll)-rt*(rt+1ll)*(rt+2ll)/6ll;
return ret;
}
inline void solve(int n){
memset(DFN,0,sizeof DFN);
memset(num,0,sizeof num);
memset(Instack,0,sizeof Instack);
Index=scc=top=0;
for(int i=1;i<=n;i++) if(!DFN[i]) Tarjan(i);
memset(indeg,0,sizeof indeg);
memset(outdeg,0,sizeof outdeg);
for(int u=1;u<=n;u++)
scce[Belong[u]].push_back(u);
for(int u=1;u<=n;u++)
for(auto enxt:e[u]){
int v=enxt.v;
if(Belong[u]!=Belong[v]){
outdeg[Belong[u]]++;
indeg[Belong[v]]++;
} else {
score[Belong[u]]+=calc(enxt.w);
}
}
}
ll find(int nowscc){
if(dp[nowscc]!=-1) return dp[nowscc];
ll ret = 0;
for(auto u:scce[nowscc])
for(auto enxt:e[u]){
int v=enxt.v;
if(Belong[u]==Belong[v]) continue;
ll w=enxt.w;
ret = max(ret,find(Belong[v])+w);
}
dp[nowscc]=(ret+score[nowscc]);
return dp[nowscc];
}
int main(){
//freopen("in.txt","r",stdin);
read(n,m);
for(int i=1;i<=m;i++){
read(x,y,z);
e[x].push_back(node(y,z));
}
solve(n);
// printf("%d
",Belong[0]);
// for(int i=1;i<=scc;i++){
// for(auto ep:scce[i])
// printf("%d ",ep);
// puts("");
// }
memset(dp,-1,sizeof dp);
int start;
read(start);
ll res = find(Belong[start]);
cout<<res;
return 0;
}