A
結論題，行著取
int n, m;
signed main(void) {
#ifdef ONLINE_JUDGE
freopen("alter.in","r",stdin);
freopen("alter.out","w",stdout);
#endif
read(n), read(m);
writeln(n / 2 + m / 2);
for (int i = 2; i <= n; i += 2)
writeln(i, ' '), writeln(1, ' '), writeln(i, ' '), writeln(m);
for (int i = 2; i <= m; i += 2)
writeln(1, ' '), writeln(i, ' '), writeln(n, ' '), writeln(i);
//fwrite(pf, 1, o1 - pf, stdout);
return 0;
}
B
按照比值排序貪心取滿。
liubw的程式碼：
const int N=1e5+10;
const db eps=1e-8;
int n,A,B;
struct node{
int a,b,g,ip;
bool operator < (const node o) const{
return a*o.b>o.a*b;
}
}p[N];
db ans[N];
void fst_IO(){
freopen("burrito.in","r",stdin);
freopen("burrito.out","w",stdout);
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
cout<<fixed<<setprecision(12);
}
int main(){
fst_IO();
cin>>n>>A>>B;
for(int i=1;i<=n;i++) cin>>p[i].g>>p[i].a>>p[i].b,p[i].ip=i;
sort(p+1,p+1+n);
db reca=0,recb=0;
for(int i=1;i<=n;i++){
db nd;
if(p[i].b>0) nd=min(p[i].g,((db)B-recb)/p[i].b);
else nd=p[i].g;
reca+=nd*p[i].a,recb+=min((db)p[i].g*p[i].b,((db)B-recb));
ans[p[i].ip]=nd;
}
if(reca<A) cout<<-1<<' '<<-1<<'\n';
else{
cout<<reca<<' '<<recb<<'\n';
for(int i=1;i<=n;i++) cout<<ans[i]<<" \n"[n==i];
}
return 0;
}
E
有限狀態自動機是唬人的本質上是隻要找到圖上的必勝環即可，dfs
int n, res[101], nxt[101][3];
int tot, ans[50001], p[50001][3];
inline void dfs(int &x, vector<int> state, int len) {
x = ++tot;
p[x][0] = p[x][1] = p[x][2] = 1;
int cnt[3] = {0, 0, 0};
for (int i : state) ++cnt[res[i]];
if (cnt[0] * cnt[1] == 0 && cnt[0] * cnt[2] == 0 && cnt[1] * cnt[2] == 0) {
ans[x] = (res[state[0]] + 1) % 3;
int trans = res[state[0]];
if (len < 2 * n) {
for (int i = 0; i < state.size(); i++)
state[i] = nxt[state[i]][ans[x]];
dfs(p[x][trans], state, len + 1);
return;
}
int u = state[0];
int T = 0;
do {
if (res[u] == 0) u = nxt[u][1];
else if (res[u] == 1) u = nxt[u][2];
else u = nxt[u][0];
++T;
} while (u != state[0]);
p[x][trans] = tot - T + 1;
return;
}
vector<int> s[3];
for (int i : state) s[res[i]].push_back(nxt[i][0]);
for (int i = 0; i < 3; i++) {
sort(s[i].begin(), s[i].end());
int l = unique(s[i].begin(), s[i].end()) - s[i].begin();
while (s[i].size() > l) s[i].pop_back();
}
ans[x] = 0;
for (int i = 0; i < 3; i++)
if (cnt[i]) dfs(p[x][i], s[i], 0);
}
signed main(void) {
freopen("epic.in", "r", stdin);
freopen("epic.out", "w", stdout);
cin >> n;
for (int i = 1; i <= n; i++) {
char c;
cin >> c;
if (c == 'R') res[i] = 0;
if (c == 'P') res[i] = 1;
if (c == 'S') res[i] = 2;
cin >> nxt[i][0] >> nxt[i][1] >> nxt[i][2];
}
vector<int> state;
for (int i = 1; i <= n; i++) state.push_back(i);
int rt;
dfs(rt, state, 0);
cout << tot << endl;
for (int i = 1; i <= tot; i++) {
if (ans[i] == 0) putchar('R');
if (ans[i] == 1) putchar('P');
if (ans[i] == 2) putchar('S');
for (int j = 0; j < 3; j++) cout << ' ' << p[i][j];
cout << endl;
}
//fwrite(pf, 1, o1 - pf, stdout);
return 0;
}
I
本質是求多少個線段能互相包含，就是LIS問題的變種。用樹狀陣列最佳化DP即可。
const int N = 2e5 + 5;
int n, p[N], f[N], g[N];
struct BIT {
int c[N];
inline void update(int x, int v) {
for (; x <= n; x += lowbit(x)) chkmax(c[x], v);
}
inline int query(int x) {
int ret = 0;
for (; x; x -= lowbit(x)) chkmax(ret, c[x]);
return ret;
}
} F, G;
signed main(void) {
#ifdef ONLINE_JUDGE
freopen("improvements.in","r",stdin);
freopen("improvements.out","w",stdout);
#endif
read(n); int ans = 0;
for (int i = 1; i <= n; i++) read(p[i]);
for (int i = 1; i <= n; i++) {
f[i] = F.query(p[i]) + 1;
g[i] = G.query(n - p[i] + 1) + 1;
chkmax(ans, f[i] + g[i] - 1);
F.update(p[i], f[i]);
G.update(n - p[i] + 1, g[i]);
}
writeln(ans);
//fwrite(pf, 1, o1 - pf, stdout);
return 0;
}
J
先算出n，然後dfs就可以了。
const int N = 2505;
char s[N]; int n, m, h[N]; vector <int> ans;
inline void dfs(int d, int lst) {
if (d > m && lst == -1 && ans.size() == n) {
for (auto u : ans) writeln(u, ' ');
exit(0);
}
if (lst == -1) {
if (!h[s[d] - '0']) h[s[d] - '0'] = 1, ans.push_back(s[d] - '0'), dfs(d + 1, -1), h[s[d] - '0'] = 0, ans.pop_back();
if (d < m) dfs(d + 1, s[d] - '0');
} else {
int val = 10 * lst + s[d] - '0';
if (val <= n && !h[val]) h[val] = 1, ans.push_back(val), dfs(d + 1, -1), h[val] = 0, ans.pop_back();
}
}
signed main(void) {
#ifdef ONLINE_JUDGE
freopen("joke.in","r",stdin);
freopen("joke.out","w",stdout);
#endif
readstr(s + 1); m = strlen(s + 1);
int s = 0;
for (n = 1; n <= 50; n++) {
if (n < 10) s++;
else s += 2;
if (s == m) break;
}
dfs(1, -1);
//fwrite(pf, 1, o1 - pf, stdout);
return 0;
}