A - Too Rich
给定每一种面值的硬币的个数，求出用最多硬币表示出所给价值。
倒着做可以用前缀和优化和剪枝。
可以$O(1)$算出当前面值减去前缀和之后最少需要几个，然后在此基础上多做3次就可以了，因为前一种硬币最多3枚就可以表示后一种硬币了。
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int dx[] = {1, 1, 5, 10, 20, 50, 100, 200, 500, 1000, 2000, 2000};
int a[20];
int T, p;
ll S[20];
int ans;
inline void upd(int &x, int y) {
if (y > x) x = y;
}
void dfs(int dep, ll val, int coins) {
//printf("%d %d %d
", dep, val, coins);
if (val < 0) {
return;
}
if (val == 0) {
upd(ans, coins);
return;
}
if (dep == 0) {
return;
}
int left = val - S[dep - 1];
if (left <= 0) {
dfs(dep - 1, val, coins);
if (a[dep] >= 1)
dfs(dep - 1, val - dx[dep], coins + 1);
} else {
int i = left / dx[dep];
if (left % dx[dep]) i++;
if (i <= a[dep])
dfs(dep - 1, val - i * dx[dep], coins + i);
else
dfs(dep - 1, val - a[dep]*dx[dep], coins + i);
i++;
if (i <= a[dep])
dfs(dep - 1, val - i * dx[dep], coins + i);
i++;
if (i <= a[dep])
dfs(dep - 1, val - i * dx[dep], coins + i);
}
}
int main() {
scanf("%d", &T);
while (T--) {
scanf("%d", &p);
for (int i = 1; i <= 10; i++)
scanf("%d", &a[i]);
for (int i = 1; i <= 10; i++)
S[i] = S[i - 1] + (ll)(a[i] * dx[i]);
ans = -1;
dfs(10, (ll)p, 0);
cout << ans << endl;
}
return 0;
}
B - Count a * b
猜了一发结论$sum_{i|n} i^2 - sum_{i|n} 1$，结果是tle，卡了$O(Tsqrt(n))$的做法，加了枚举的优化。
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int N = 1e5;
bool p[100005];
int prime[100005],top;
struct node{
int x,y;
}ort[10005];
void init(){
memset(p, 0, sizeof(p));
top = 0;
for (int i = 2; i <= N; ++i) {
if (!p[i]) {
prime[++top] = i;
}
for (int j = 1; j <= top; ++j) {
if (i * prime[j] > N)
break;
p[i * prime[j]] = 1;
if (i % prime[j] == 0)
break;
}
}
}
ull r1,r2;
int ctop;
void dfs(int dep,int now){
if(dep==ctop+1){
r1+=(ull)now*now;
return;
}
int t=1;
for(int i=0;i<=ort[dep].y;i++){
dfs(dep+1,now*t);
t*=ort[dep].x;
}
}
int main() {
//freopen("in.txt","r",stdin);
init();
int n, T;
cin >> T;
while (T--) {
scanf("%d", &n);
int tmp=n;
// ull res1 = 0, res2 = 0;
// for (int i = 1; i * i <= n; i++) {
// if (n % i == 0) {
// res1 += (ull)i * i;
// res2 ++;
// if (i * i != n) {
// int t = n / i;
// res1 += (ull)t * t;
// res2 ++;
// }
// }
// }
// res1 -= (ull)n * res2;
// printf("%llu
", res1);
ctop=0;
for(int i=1;i<=top && prime[i]*prime[i]<=n;i++){
if(n%prime[i]==0){
ctop++;
ort[ctop].x=prime[i];
ort[ctop].y=0;
while(n%prime[i]==0){
ort[ctop].y++;
n/=prime[i];
}
}
}
if(n!=1){
ctop++;
ort[ctop].x=n;
ort[ctop].y=1;
}
r1=0;r2=1;
for(int i=1;i<=ctop;i++)
r2=r2*(ull)(ort[i].y+1);
r2=r2*(ull)tmp;
dfs(1,1);
cout<<r1-r2<<endl;
}
return 0;
}
F - Almost Sorted Array
判断是不是almost sorted array。
跟着跑LIS，如果栈高比当前i小太多直接就是不对的了。
所以复杂度还是$O(n)$。
#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 5;
int T, n, k, a[maxn], k1, k2, ans[maxn], len, b[maxn], flag;
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);}
int main() {
//freopen("in.txt", "r", stdin);
read(T);
while (T--) {
flag = 0;
memset(ans, 0, sizeof(ans));
k1 = k2 = 0;
scanf("%d", &n);
for (int i = 1; i <= n; i++)
read(a[i]);
ans[1] = a[1];
len = 1;
for (int i = 2; i <= n; i++) {
if (a[i] >= ans[len])
ans[++len] = a[i];
else {
int pos = upper_bound(ans + 1, ans + len + 1, a[i]) - ans;
ans[pos] = a[i];
}
if(i-len>5) break;
}
if (len >= n - 1) {
puts("YES");
continue;
}
for (int i = 1; i <= n; i++)
b[i] = a[n - i + 1];
ans[1] = b[1];
len = 1;
for (int i = 2; i <= n; i++) {
if (b[i] >= ans[len])
ans[++len] = b[i];
else {
int pos = upper_bound(ans + 1, ans + len + 1, b[i]) - ans;
ans[pos] = b[i];
}
if(i-len>5) break;
}
if (len >= n - 1) puts("YES");
else puts("NO");
}
return 0;
}
G - Dancing Stars on Me
每个点只有两个最近点，并且最近的距离相等。
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
double eps = 1e-8;
inline int dcmp(double x) {
if (x > eps) return 1;
if (x < -eps) return -1;
return 0;
}
struct node {
int x, y;
} info[1005];
int cnt[1005];
double getdist(int i, int j) {
double rx = info[i].x - info[j].x;
double ry = info[i].y - info[j].y;
double ret = rx * rx + ry * ry;
return ret;
}
int main() {
//freopen("in.txt","r",stdin);
int T,n;
cin >> T;
while (T--) {
scanf("%d", &n);
for (int i = 1; i <= n; i++)
scanf("%d%d", &info[i].x, &info[i].y);
double cmp = 1e15;
for (int i = 2; i <= n; i++) {
cmp = min(cmp, getdist(1, i));
}
bool flag=1;
memset(cnt,0,sizeof cnt);
for(int i=1;i<=n;i++){
for(int j=1;j<=n;j++){
if(i==j) continue;
if(dcmp(getdist(i,j)-cmp)==0) cnt[i]++;
if(dcmp(getdist(i,j)-cmp)<0){
flag=0;
break;
}
}
}
for(int i=1;i<=n;i++)
if(cnt[i]!=2){
flag=0;
break;
}
puts(flag?"YES":"NO");
}
return 0;
}
H - Partial Tree
2n-2个度分配给n个点。
原本是一个二维消费的背包。
有一个特殊的条件就是每个点至少一个度，那么可以直接先把这些度分配给n个点。
然后就是n-2个度任意分配的背包问题，把得到的价值修改成a[i+1]-a[1]就行了。
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
int f[10005];
int dp[4400],res;
int main() {
//freopen("in.txt","r",stdin);
int T,n;
cin >> T;
while (T--) {
scanf("%d",&n);
for(int i=1;i<n;i++)
scanf("%d",&f[i]);
res = (ll)f[1]*n;
//cout<<res<<endl;
for(int i=2;i<n;i++)
f[i]-=f[1];
for(int i=0;i<n;i++)
dp[i]=-INT_MAX;
dp[0]=0;
for(int i=1;i<n-1;i++)
for(int j=i;j<=n-2;j++){
dp[j]=max(dp[j],dp[j-i]+f[i+1]);
}
// for(int i=1;i<=n-2;i++)
// printf("%d ",dp[i]);
// puts("");
cout<<dp[n-2]+res<<endl;
}
return 0;
}
J - Chip Factory
居然可以暴力过。
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
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);}
int a[1005];
int main() {
//freopen("in.txt","r",stdin);
int T, n;
read(T);
while (T--) {
read(n);
for (int i = 1; i <= n; i++)
read(a[i]);
int Mx = 0;
for (int i = 1; i < n; i++)
for (int j = i + 1; j < n; j++)
for (int k = j + 1; k <= n; k++) {
Mx = max(Mx, (a[i] + a[j])^a[k]);
Mx = max(Mx, (a[i] + a[k])^a[j]);
Mx = max(Mx, (a[k] + a[j])^a[i]);
}
printf("%d
",Mx);
}
return 0;
}
然后$O(n^2 log n)$，枚举$S_i$,$S_j$从字典树删除,贪心查询$S_k$。
#include <bits/stdc++.h>
using namespace std;
const int maxn = 10000+5;
typedef long long ll;
ll bin[35];
int n,m;
ll a[maxn],x;
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 trie{
int cnt;
int ch[maxn*35][2];
ll val[maxn*35];
int ts[maxn*35];
inline void init(){
cnt=1;
memset(ch[0],0,sizeof ch[0]);
ts[0]=0;
//memset(ts,0,sizeof ts);
}
inline void insert(ll a){
int u=0;
for(int i=32;i>=0;i--){
ll t=a&bin[i];t>>=i;
if(!ch[u][t]){
memset(ch[cnt],0,sizeof ch[cnt]);
val[cnt]=0;
ts[cnt]=0;
ch[u][t]=cnt++;
}
u=ch[u][t];
ts[u]++;
}
val[u]=a;
}
inline void del(ll a){
int u=0;
for(int i=32;i>=0;i--){
ll t=a&bin[i];t>>=i;
u=ch[u][t];
ts[u]--;
}
}
inline ll query(ll a){
int u=0;
for(int i=32;i>=0;i--){
ll t=a&bin[i];t>>=i;
if(ch[u][t^1]&&ts[ch[u][t^1]]) u=ch[u][t^1];
else u=ch[u][t];
}
return val[u]^a;
}
}Trie;
int main(){
//freopen("in.txt","r",stdin);
bin[0]=1;for(ll i=1;i<=35;i++) bin[i]=bin[i-1]*2ll;
int T,cas=1;
for(cin>>T,cas=1;cas<=T;cas++){
scanf("%d",&n);
Trie.init();
for(int i=1;i<=n;i++){
read(a[i]);
Trie.insert(a[i]);
}
ll mx=0;
for(int i=1;i<=n;i++)
for(int j=i+1;j<=n;j++){
Trie.del(a[i]);Trie.del(a[j]);
mx=max(mx,Trie.query(a[i]+a[j]));
Trie.insert(a[i]);Trie.insert(a[j]);
}
printf("%lld
",mx);
}
return 0;
}
L - House Building
#include <bits/stdc++.h>
#define rep(i,x,y) for(int i=x;i<=y;i++)
#define per(i,x,y) for(int i=x;i>=y;i--)
#define debug(a) cerr<<#a<<"=="<<a<<endl
#define iout(x) printf("%d
",x)
#define lout(x) printf("%lld
",x)
#define iin(x) scanf("%d",&x)
#define lin(x) scanf("%lld",&x)
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
using namespace std;
typedef vector<int> VI;
typedef long long ll;
typedef pair<int, int> PII;
const int mod=1e9+7;
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);}
template<typename A,typename B> inline A fexp(A x,B p){A ans=1;for(;p;p>>=1,x=1LL*x*x%mod)if(p&1)ans=1LL*ans*x%mod;return ans;}
template<typename A,typename B> inline A fexp(A x,B p,A mo){A ans=1;for(;p;p>>=1,x=1LL*x*x%mo)if(p&1)ans=1LL*ans*x%mo;return ans;}
//head
int t;
int a[100][100];
int n,m;
int main() {
//freopen("in.txt","r",stdin);
read(t);
while(t--) {
memset(a,0,sizeof(a));
scanf("%d%d",&n,&m);
for(int i=1;i<=n;i++) {
for(int j=1;j<=m;j++) {
scanf("%d",&a[i][j]);
}
}
int ans=0;
rep(i,1,n) {
rep(j,1,m) {
if(a[i][j]!=0) ans++;
if(a[i][j]>a[i-1][j]) ans+=a[i][j]-a[i-1][j];
if(a[i][j]>a[i+1][j]) ans+=a[i][j]-a[i+1][j];
if(a[i][j]>a[i][j-1]) ans+=a[i][j]-a[i][j-1];
if(a[i][j]>a[i][j+1]) ans+=a[i][j]-a[i][j+1];
}
}
printf("%d
",ans);
}
return 0;
}