Codeforces Round 929 (Div. 3)
A. Turtle Puzzle: Rearrange and Negate
题意:可以对整数数组进行两个操作,一是随意重新排列或保持不变,二是选择连续子段元素符号倒转,求可能最大的所有元素和。
思路:所有元素的绝对值和。
AC code:
c++
void solve() {
int sum = 0;
cin >> n;
for (int i = 0; i < n; i ++) {
int x; cin >> x;
sum += abs(x);
}
cout << sum << endl;
}
B. Turtle Math: Fast Three Task
题意:给出整数数组a,可以选择任意两种操作,一是选择一个元素从数组a中删除,二是选择一个元素数值增加1;最少需要多少次操作可以使得数组元素之和能被3整除。
思路:
- 元素和本来就能被3整除,0次操作;
- 元素和mod3=2,1次添加操作;
- 元素和mod3=1,若存在删除数组中的一个元素使得元素和被3整除,则1次操作,否则2次添加操作。
AC code:
c++
void solve() {
cin >> n;
int sum = 0;
for (int i = 0; i < n; i ++) {
cin >> a[i];
sum += a[i];
}
if (sum % 3 == 0) {
cout << 0 << endl;
return;
}
int u = sum % 3;
if (u == 2) {
cout << 1 << endl;
return;
}
for (int i = 0; i < n; i ++) {
if ((sum - a[i]) % 3 == 0) {
cout << 1 << endl;
return;
}
}
cout << 2 << endl;
}
C. Turtle Fingers: Count the Values of k
题意:给出正整数a, b, l,若存在 k k k, x x x, y y y 使得 l = k ⋅ a x ⋅ b y l = k \cdot a^x \cdot b^y l=k⋅ax⋅by,则k有多少种不同的可能。
思路:暴力枚举x,y,注意处理枚举边界,即处理出x和y的最大可能再进行枚举。
AC code:
c++
int tmp(int l, int t) {
int pos = 0;
while (l % t == 0) {
l /= t;
pos ++;
}
return pos;
}
void solve() {
int a, b, l; cin >> a >> b >> l;
int cnt = 0;
map<int, int> mp;
int A = tmp(l, a);
int B = tmp(l, b);
//cout << A << " " << B <<"+++" << endl;
for (int i = 0; i <= A; i ++) {
for (int j = 0; j <= B; j ++) {
int t = pow(a, i) * pow(b, j);
if (l % t == 0) {
mp[l / t]++;
}
}
}
cout << mp.size() << endl;
}
D. Turtle Tenacity: Continual Mods
题意:给定正整数数组a,重排后得到数组b,是否可能存在 b 1 m o d b 2 m o d ... m o d b n ≠ 0 b_1 \bmod b_2 \bmod \ldots \bmod b_n \neq 0 b1modb2mod...modbn=0.
思路:找出所有元素的最小公倍数gcd,若存在题述情况,则gcd出现次数一定小于2,否则mod过程中会出现0.
AC code:
c++
int gcd(int a, int b) {
if (b) while ((a%=b) && (b%=a));
return a + b;
}
void solve() {
cin >> n;
int now = -1;
for (int i = 1; i <= n; i ++) {
cin >> a[i];
if (now == -1) {
now = a[i];
} else {
now = gcd(now, a[i]);
}
}
int cnt = 0;
for (int i = 1; i <= n; i ++) {
if (a[i] == now) cnt ++;
}
if (cnt < 2) {
cout << "YES" << endl;
} else {
cout << "NO" << endl;
}
}
E. Turtle vs. Rabbit Race: Optimal Trainings
题意:有n条跑道,每条跑道有 a i a_i ai个部分,给定一个正整数u,每完成一个部分成绩提高u,然后u-1,给出q组l和u,找出一个下标r,使得[l,r]的跑道成绩最高。
思路:二分找到最接近u的区间即可,注意二分的上边界和下边界都有可能成为答案,找到上下最接近u的边界,然后取其中离u最近的部分。
AC code:
c++
void solve() {
cin >> n;
for (int i = 0; i <= n; i ++) {
sum[i] = 0;
}
for (int i = 1; i <= n; i ++) {
cin >> a[i];
sum[i] = sum[i - 1] + a[i];
}
cin >> q;
while (q --) {
int l, u; cin >> l >> u;
int L = l, R = n;
while (L < R) {
int mid = L + R + 1 >> 1;
if (sum[mid] - sum[l - 1] <= u) L = mid;
else R = mid - 1;
}
int pos1 = L;
L = l, R = n;
while (L < R) {
int mid = L + R >> 1;
if (sum[mid] - sum[l - 1] >= u) R = mid;
else L = mid + 1;
}
int pos2 = R;
//cout << pos1 << " " << pos2 << "++++" << endl;
int now1 = abs(sum[pos1] - sum[l - 1] - u), now2 = abs(sum[pos2] - sum[l - 1] - u);
if (now1 >= now2) cout << pos2 << " ";
else cout << pos1 << " ";
} cout << endl;
}