2025 ICPC Gran Premio de Mexico 3ra Fecha

A题

签到

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 1e6+5;
ll n;
ll a[N];
int main() {ios_base::sync_with_stdio(false),cin.tie(0),cout.tie(0);cin>>n;ll sum = 0;for(int i=0;i<n;i++) {cin>>a[i];sum+=a[i];}ll t = n;while (t >= 1) {if (sum % t == 0) {cout<<n - t <<'\n';return 0;}t--;}return 0;
}

二、K题

分类讨论

#include<bits/stdc++.h>
#define int long long
using namespace std;void solve() {int a,b,c,d;cin>>a>>b>>c>>d;if (d <= a) {cout<<"0\n";return;}if (d > a && d <= b) {int ans = d - max(a,c);cout<<ans<<"\n";return;}if (d > b) {if (c <= a) {cout<<b - a <<'\n'; return;}if (c > a && c <= b){cout << b - c <<'\n'; return;}if (c > b){cout << "0\n";return;}}
}
signed main() {ios_base::sync_with_stdio(false),cin.tie(0),cout.tie(0);int T;cin>>T;while(T--) {solve();}return 0;
}

三、B题

贡献法
结论就是i * (n - i + 1) * a[i]是每个值的贡献。

#include<bits/stdc++.h>
const int N = 1e5;
#define int long long
using namespace std;
int a[N];
void solve() {int n;cin>>n;for(int i=1;i<=n;i++) {cin>>a[i];}int ans = 0;for(int i=1;i<=n;i++) {ans += i * (n - i + 1) * a[i];}cout<<ans<<'\n';
}
signed main() {ios_base::sync_with_stdio(false),cin.tie(0),cout.tie(0);int T = 1;// cin>>T;while(T--) {solve();}return 0;
}

四、J题

线段树 + 尼姆博弈

#include <bits/stdc++.h>
using namespace std;const int MAX_N = 1e6 + 5;
long long a[MAX_N];
long long sum_xor[MAX_N << 2];
//建树
void build(int l, int r, int rt) {if (l == r) {sum_xor[rt] = a[l];return;}int mid = (l + r) >> 1;build(l, mid, rt << 1);build(mid + 1, r, rt << 1 | 1);sum_xor[rt] = sum_xor[rt << 1] ^ sum_xor[rt << 1 | 1];//up函数
}void update(int k, long long x, int l, int r, int rt) {if (l == r) {//注意：这里是叶子节点异或值等于本身，加上后仍然是其异或值sum_xor[rt] += x;return;}int mid = (l + r) >> 1;if (k <= mid) {update(k, x, l, mid, rt << 1);} else {update(k, x, mid + 1, r, rt << 1 | 1);}sum_xor[rt] = sum_xor[rt << 1] ^ sum_xor[rt << 1 | 1];//up函数
}
//范围查询
long long query(int L, int R, int l, int r, int rt) {if (L <= l && r <= R) {return sum_xor[rt];}int mid = (l + r) >> 1;long long res = 0;if (L <= mid) {res ^= query(L, R, l, mid, rt << 1);}if (R > mid) {res ^= query(L, R, mid + 1, r, rt << 1 | 1);}return res;
}int main() {ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);int n, q;scanf("%d%d", &n, &q);for (int i = 1; i <= n; ++i) {scanf("%lld", &a[i]);}build(1, n, 1);char op[2];while (q--) {scanf("%s", op);if (op[0] == 'P') {int l, r;scanf("%d%d", &l, &r);long long xor_sum = query(l, r, 1, n, 1);if (xor_sum == 0) {printf("JUAN\n");} else {printf("FRANK\n");}} else if (op[0] == 'R') {int k;long long x;scanf("%d%lld", &k, &x);update(k, x, 1, n, 1);}}return 0;
}

五、G题

经典精度问题，一定要想到要将除法转换为乘法，转换后因为数值较大，可以转为log，乘变加。

#include<bits/stdc++.h>
#define int long long
using namespace std;
typedef long double ld;
void solve() {int h1, h2, b;cin >> h1 >> h2 >> b;ld t = logl(ld (h1) / h2);ld r = logl(ld (b)) - logl(ld (b-1));ld ans = t / r;int res = int32_t(ceil(ans));cout << res << '\n';
}signed main() {ios_base::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);int T;cin >> T;while (T--) {solve();}return 0;
}