今天這場沒準備打的,但是lbw打,我就倒開看看題目了。非常意外的是這場只要做到E就可以100名,而我感覺除了F都是垃圾題。
D. Minimize the Difference
這種求極差的題目很常規,由他那個操作可以看出單調性。故可以兩遍二分分別求最小值的最大值和最大值的最小值然後做差,也可以用單調棧去做。
const int N = 2e5 + 5;
int T, n; ll a[N];
inline bool check(ll x) {
ll sum = 0;
for (int i = 1; i <= n; i++) {
if (a[i] > x) sum += a[i] - x;
if (a[i] < x) {
if (x - a[i] > sum) return false;
sum -= x - a[i];
}
} return true;
}
inline bool checkII(ll x) {
ll sum = 0;
for (int i = 1; i <= n; i++) {
if (a[i] > x) sum += a[i] - x;
if (a[i] < x) sum -= x - a[i], sum = max(sum, 0ll);
}
if (sum > 0) return false;
return true;
}
signed main(void) {
for (read(T); T; T--) {
read(n); ll mx = 0, mn = 1ll << 62;
for (int i = 1; i <= n; i++)
read(a[i]), chkmax(mx, a[i]), chkmin(mn, a[i]);
ll l = mn, r = mx, ans1 = l;
while (l <= r) {
ll mid = l + r >> 1;
if (check(mid)) ans1 = mid, l = mid + 1;
else r = mid - 1;
}
l = mn, r = mx; ll ans2 = r;
while (l <= r) {
ll mid = l + r >> 1;
if (checkII(mid)) ans2 = mid, r = mid - 1;
else l = mid + 1;
}
// writeln(ans1, ' '); writeln(ans2);
writeln(ans2 - ans1);
}
//fwrite(pf, 1, o1 - pf, stdout);
return 0;
}
E. Prefix GCD
我們知道在gcd變小的過程中至少要除以二,則gcd收斂的次數是log級別的。故可以貪心找每次填入後字首gcd最小的數,時間複雜度 \(O(n\log(A))\)
inline int gcd(int a, int b) {
return b == 0 ? a : gcd(b, a % b);
}
const int N = 1e5 + 5;
int T, n, a[N];
signed main(void) {
for (read(T); T; T--) {
read(n);
for (int i = 1; i <= n; i++) read(a[i]);
if (n == 1) { writeln(a[1]); continue; }
sort(a + 1, a + n + 1);
int d = gcd(a[1], a[2]);
for (int i = 3; i <= n; i++) d = gcd(d, a[i]);
int D = a[1] / d, j = 1; ll ans = D;
while (D > 1 && j < n) {
int M = INT_MAX, k = j + 1;
for (int i = j + 1; i <= n; i++)
if (M > gcd(a[i] / d, D)) {
M = gcd(a[i] / d, D);
k = i;
}
ans += M; swap(a[j + 1], a[k]); D = M; j++;
// writeln(D, ' '); writeln(j);
}
ans += n - j;
writeln(1ll * ans * d);
}
//fwrite(pf, 1, o1 - pf, stdout);
return 0;
}