2022“杭电杯”中国大学生算法设计超级联赛（10） 题解
C. Wavy Tree
发现修改次数和相邻两数的相对大小有关，所以可先求出差分数组。
分两种情况考虑：① 奇数位置为波峰 ② 偶数位置为波峰。
以情况 ① 为例，若奇数位置差分后值小于等于0则不合法，需要修改至1；若偶数位置差分后值大于等于0则不合法，需要修改至-1。
情况 ② 同理。
#include<bits/stdc++.h>
#define int long long
using namespace std;
const int MAXN = 1e6 + 5;
int n, a[MAXN], b[MAXN], d[MAXN];
int Solve(int x) {
int res = 0;
for(int i = 1; i <= n; i++) a[i] = d[i];
//for(int i = 1; i <= n; i++) cout << a[i] << " "; cout << " a\n";
for(int i = 2; i <= n; i++) {
if(i % 2 == x) {
if(a[i] <= 0) {
int c = 1 - a[i];
res += c;
a[i + 1] -= c;
a[i] = 1;
}
} else {
if(a[i] >= 0) {
int c = a[i] + 1;
res += c;
a[i + 1] += c;
a[i] = -1;
}
}
}
return res;
}
void Solve() {
cin >> n;
for(int i = 1; i <= n; i++) cin >> b[i];
for(int i = 1; i <= n; i++) d[i] = b[i] - b[i - 1];
int ans = min( Solve(0), Solve(1) );
cout << ans << "\n";
}
signed main()
{
ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
int T; cin >> T;
while(T--) Solve();
return 0;
}
D. Average Replacement
猜了两个结论：① 同一个连通块的所有点最后值相等；② 同一个连通块内 \(\sum (deg_i+1)a_i\) 为定值，其中 \(deg_i\) 为 \(i\) 的度数。
最后猜了每个连通块中点的值则为 \(\frac{\sum(deg_i+1)a_i}{\sum(deg_i+1)}\)。
#pragma GCC optimize(2)
#include<bits/stdc++.h>
using namespace std;
namespace fastIO{
#define BUF_SIZE 100000
#define OUT_SIZE 100000
#define ll long long
bool IOerror=0;
int qr;
char obuf[OUT_SIZE],*oS=obuf,*oT=oS+OUT_SIZE-1,qu[55];
inline void flush(){
fwrite(obuf,1,oS-obuf,stdout);
oS=obuf;
}
inline char nc(){
static char buf[BUF_SIZE],*p1=buf+BUF_SIZE,*pend=buf+BUF_SIZE;
if (p1==pend){
p1=buf;pend=buf+fread(buf,1,BUF_SIZE,stdin);
if(pend==p1){IOerror=1;return -1;}
}
return *p1++;
}
inline bool blank(char ch){return ch==' '||ch=='\n'||ch=='\r'||ch=='\t';}
inline void read(int &x){
bool sign=0; char ch=nc(); x=0;
for(;blank(ch);ch=nc());
if(IOerror)return;
if(ch=='-')sign=1,ch=nc();
for(;ch>='0'&&ch<='9';ch=nc())x=x*10+ch-'0';
if(sign)x=-x;
}
inline void read(ll &x){
bool sign=0;char ch=nc();x=0;
for(;blank(ch);ch=nc());
if(IOerror)return;
if(ch=='-')sign=1,ch=nc();
for(;ch>='0'&&ch<='9';ch=nc())x=x*10+ch-'0';
if(sign)x=-x;
}
inline void read(float &x){
x=0;char ch=nc();int w=1,cnt=1,n=0;
for(;blank(ch);ch=nc());
if(IOerror)return;
if(ch=='-') w=-w,ch=nc();
for(;ch>='0'&&ch<='9';ch=nc())x=x*10+ch-'0';
if(ch=='.'){ch=nc();for(;ch>='0'&&ch<='9';ch=nc())n=n*10+ch-'0',cnt*=10;}
x=1.0*w*(x+1.0*n/cnt);
}
inline void read(double &x){
x=0;char ch=nc();int w=1,cnt=1,n=0;
for(;blank(ch);ch=nc());
if(IOerror)return;
if(ch=='-') w=-w,ch=nc();
for(;ch>='0'&&ch<='9';ch=nc())x=x*10+ch-'0';
if(ch=='.'){ch=nc();for(;ch>='0'&&ch<='9';ch=nc())n=n*10+ch-'0',cnt*=10;}
x=1.0*w*(x+1.0*n/cnt);
}
inline void read(char &x){
char ch=nc();
for(;blank(ch);ch=nc());
if(IOerror)return;
x=ch;
}
inline void write(char x){
*oS++=x;
if(oS==oT)flush();
}
template <class I>
inline void write(I x){
if(!x) write('0');if(x < 0) write('-'),x=-x;
while (x) qu[++qr]=x%10+'0',x/=10;
while (qr) write(qu[qr--]);
}
struct Flusher_{~Flusher_(){flush();}}io_flusher_;
#undef ll
#undef OUT_SIZE
#undef BUF_SIZE
};
using namespace fastIO;
const int MAXN = 1e5 + 5;
int n, m, a[MAXN], f[MAXN], u[MAXN], v[MAXN];
long long tot[MAXN], sze[MAXN];
int find(int x) { return x == f[x] ? x : f[x] = find(f[x]); }
void merge(int u, int v) {
int fu = find(u), fv = find(v);
if(fu == fv) return ;
if(fu > fv) { swap(u, v); swap(fu, fv); }
f[fv] = fu;
sze[fu] += sze[fv], sze[fv] = 0;
tot[fu] += tot[fv], tot[fv] = 0;
}
void Solve() {
read(n); read(m);
for(int i = 1; i <= n; ++i) {
f[i] = i;
tot[i] = 0;
sze[i] = 1;
}
for(int i = 1; i <= n; ++i) read(a[i]);
for(int i = 1; i <= m; ++i) {
read(u[i]); read(v[i]);
sze[ u[i] ]++; sze[ v[i] ]++;
}
//for(int i = 1; i <= n; i++) write(sze[i]), write(' '); write('\n');
for(int i = 1; i <= n; ++i) tot[i] = 1ll * sze[i] * a[i];
for(int i = 1; i <= m; i++) merge(u[i], v[i]);
//for(int i = 1; i <= n; i++) write(tot[i]), write(' '); write('\n');
//for(int i = 1; i <= n; i++) write(sze[i]), write(' '); write('\n');
for(int i = 1; i <= n; ++i) {
int fi = find(i);
double ans = 1.0 * tot[fi] / (1.0 * sze[fi]);
printf("%.6lf\n", ans);
/*
int d1 = tot[fi] / sze[fi];
int d2 = 1ll * ((tot[fi] % sze[fi]) * 1000000ll) / sze[fi];
int cnt = 0, cur = d2;
while(cur) cur /= 10, ++cnt;
//double ans = 1.0 * tot[fi] / (1.0 * sze[fi]);
//printf("%.6lf", ans);
write(d1);
write('.');
for(int j = cnt + 1; j <= 6; j++) write('0');
if(d2) write (d2);
write('\n');
*/
}
}
signed main()
{
int T; read(T);
while(T--) Solve();
return 0;
}
I. Painting Game
设当前最后一个被涂黑的位置为 \(pos\)，对于 \(Alice\)，其下一个所涂位置一定为 \(pos+3\)；对于 \(Bob\)，其下一个所涂位置一定为 \(pos+4\)，最后阶段再返回来涂 \(pos+2\)。讨论一下边界，中间过程可以直接通过每次跳7步的方式加速，具体看代码。
#pragma GCC optimize(2)
#include<bits/stdc++.h>
using namespace std;
const int MAXN = 1e5 + 5;
int n;
string S;
void Solve() {
cin >> n >> S;
int pos = 0, cnt = 0, st;
if(n == 1 || n == 2) return void(cout << "1\n");
if(S[0] == 'B') {
st = 1;
pos = 3;
cnt += 2;
st = 0;
int k = (n - pos) / 7;
cnt += 3 * k;
pos += 7 * k;
while(pos <= n) {
if(pos + 3 > n) break;
pos += 3, cnt += 1;
st = 1;
if(pos + 4 > n) break;
pos += 4, cnt += 2;
st = 0;
}
if(!pos) {
if(pos + 1 == n) cnt++;
if(pos + 2 == n) cnt++;
if(pos + 3 == n) {
cnt++;
if(st) cnt++;
}
if(pos + 4 == n) {
cnt += 2;
}
} else {
if(pos + 1 == n) ;
if(pos + 2 == n) cnt++;
if(pos + 3 == n) cnt++;
if(pos + 4 == n) {
cnt++;
if(st) cnt++;
}
}
} else {
st = 0;
pos = 2;
cnt++;
st = 1;
int k = (n - pos) / 7;
cnt += 3 * k;
pos += 7 * k;
while(pos <= n) {
if(pos + 4 > n) break;
pos += 4, cnt += 2;
st = 0;
if(pos + 3 > n) break;
pos += 3, cnt += 1;
st = 1;
}
if(!pos) {
if(pos + 1 == n) cnt++;
if(pos + 2 == n) cnt++;
if(pos + 3 == n) {
cnt++;
if(st) cnt++;
}
if(pos + 4 == n) {
cnt += 2;
}
} else {
if(pos + 1 == n) ;
if(pos + 2 == n) cnt++;
if(pos + 3 == n) cnt++;
if(pos + 4 == n) {
cnt++;
if(st) cnt++;
}
}
}
cout << cnt << "\n";
}
signed main()
{
ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
int T; cin >> T;
while(T--) Solve();
return 0;
}
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