牛客小白月赛53
牛客小白月赛53
第二次打妞妞月赛
三题下班，呜呜呜我好菜orz
https://ac.nowcoder.com/acm/contest/11230
A. Raining
按照题意模拟即可
#include <bits/stdc++.h>
using namespace std;
int main () {
int a, b, c, d, x;
cin >> a >> b >> c >> d >> x;
a = max (x-a, 0), d = max (x-d, 0), c = max (x-c, 0), b = max (x-b, 0);
cout << a << ' ' << b << ' ' << c << ' ' << d << endl;
}
B. Kissing
结论题, 先对通项化简:\(i^2-((i+1)^2-4i)=i^2-(i-2)^2=2i-1\)
然后是等差数列求和公式,得\(n^2\)
会爆long long, 所以要随时取模
#include <bits/stdc++.h>
using namespace std;
const int mod = 998244353;
typedef long long ll;
int main () {
ll n;
cin >> n;
cout << ((n%mod)*(n%mod))%mod;
}
//2*i-1
C. Missing
按照题意模拟+结构体排序输出
(一开始前面写的while(n--),结果一直没更新答案,看了半天,蠢死了呜呜呜)
#include <bits/stdc++.h>
using namespace std;
typedef pair<int, string> psi;
int main () {
string s;
int n;
cin >> s >> n;
vector <psi> v;
vector <string> ss;
for (int i = 0; i < n; i ++) {
string t;
cin >> t;
ss.push_back (t);
}
sort (ss.begin (), ss.end ());
//for (auto i : ss) cout << i << ' '; cout << endl;
for (int i = 0; i < n; i ++) {
string t = ss[i];
if (t.size() != s.size()) v.push_back ({0, t});
else {
int cnt = 0;
for (int i = 0; i < s.size (); i ++)
if (s[i] == t[i]) cnt ++;
v.push_back ({cnt, t});
}
}
//for (auto i : v) cout << i.first << ' ' << ss[i.second] << endl; cout << endl;
sort (v.begin (), v.end (), greater<psi>());
int maxn = v[0].first;
//cout << maxn << endl;
ss.clear();
for (auto i : v) {
if (i.first != maxn) break;
ss.push_back (i.second);
}
sort (ss.begin (), ss.end());
for (auto i : ss) cout << i << endl;
}
//不除
//如果「相似度」都为 0，请仍然按字典序输出所有字符串
D. Breezing
显然,a[i]取1或b[i]的时候是最优的
那么就可以dp(我连最简单的dp都不会)
定义一个二维dp数组f[N][2]:
f[i][0]表示以为i结尾,当前位置选取a[i]=1,时的最大可爱值
f[i][1]表示以为i结尾,当前位置选取a[i]=b[i],时的最大可爱值
正解:
#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 5;
int f[N][2], a[N];
int main () {
int n;
cin >> n;
for (int i = 1; i <= n; i ++) cin >> a[i];
for (int i = 2; i <= n; i ++) {
f[i][0] = max (f[i-1][0], f[i-1][1] + abs (a[i-1] - 1));
f[i][1] = max (f[i-1][0] + abs (a[i] - 1), f[i-1][1] + abs (a[i-1] - a[i]));
}
cout << max(f[n][0], f[n][1]);
}
原来的错误写法:贪心,一大一小
贪心有风险,要慎重www
#include <bits/stdc++.h>
#define int long long
using namespace std;
signed main () {
int n;
cin >> n;
vector <int> a(n+1);
for (int i = 1; i <= n; i ++) cin >> a[i];
int ans1 = 0, ans2 = 0;
//偶数
if (n % 2 == 0) {
for (int i = 2; i <= n; i += 2)
ans1 += (a[i] - 1) * 2, ans2 += (a[i-1] - 1) * 2;
ans1 -= (a[n] - 1), ans2 -= (a[1] - 1);
}
else {
for (int i = 2; i <= n; i += 2)
ans1 += (a[i] - 1) * 2, ans2 += (a[i-1] - 1) * 2;
ans2 -= (a[1] - 1), ans2 -= (a[n] - 1);
if (n == 3) ans2 = a[1] - 1 + a[3] - 1;
}
//cout << ans1 << ' ' << ans2 << endl;
cout << max(max (ans1, ans2), 0ll);
}
//1 b[i] 1 b[i]....
//b[i] 1 b[i] 1...
E. Calling
终于知道问题出在哪里了，res[2]更新的时候不太对，这样写就没问题了：（感谢Hytidel佬的指正）
（因为在更新的时候，首先消耗整的，再来考虑余下的）所以要先统一减cnt*9
#include <bits/stdc++.h>
#define int long long
using namespace std;
void solve () {
int n;
int num[7];
cin >> n;
for (int i = 1; i <= 6; i ++) cin >> num[i];
// if ((n - num[4] - num[5] - num[6]) < 0) {cout << "No\n"; return ;}
int res[7] = {0, 36*n, 9*n, 4*n, n, n, n};
if (res[6] < num[6]) {cout << "No\n"; return ;}
res[5] -= num[6], res[4] -= num[6], res[3] -= num[6]*4, res[2] -= num[6]*9, res[1] -= num[6]*36;
if (res[5] < num[5]) {cout << "No\n"; return ;}
res[4] -= num[5], res[3] -= num[5]*4, res[2] -= num[5]*9, res[1] -= num[5]*25;
if (res[4] < num[4]) {cout << "No\n"; return ;}
res[3] -= num[4]*4, res[2] -= num[4]*4, res[1] -= num[4]*16;
if (res[3] < num[3]) {cout << "No\n"; return ;}
res[1] -= num[3]*9;
int mod = num[3] % 4, cnt = num[3]/4;
res[2] -= cnt * 9;
if (mod == 3) res[2] -= 8;
else if (mod == 2) res[2] -= 6;
else if (mod == 1) res[2] -= 4;
if (res[2] < num[2]) {cout << "No\n"; return ;}
res[1] -= num[2]*4;
if (res[1] < num[1]) {cout << "No\n"; return ;}
cout << "Yes\n";
}
signed main () {
int t;
cin >> t;
while (t --) solve ();
}
原来的错误写法:嗯模拟
#include <bits/stdc++.h>
#define int long long
using namespace std;
void solve () {
int n;
int num[7];
cin >> n;
for (int i = 1; i <= 6; i ++) cin >> num[i];
// if ((n - num[4] - num[5] - num[6]) < 0) {cout << "No\n"; return ;}
int res[7] = {0, 36*n, 9*n, 4*n, n, n, n};
if (res[6] < num[6]) {cout << "No\n"; return ;}
res[5] -= num[6], res[4] -= num[6], res[3] -= num[6]*4, res[2] -= num[6]*9, res[1] -= num[6]*36;
if (res[5] < num[5]) {cout << "No\n"; return ;}
res[4] -= num[5], res[3] -= num[5]*4, res[2] -= num[5]*9, res[1] -= num[5]*25;
if (res[4] < num[4]) {cout << "No\n"; return ;}
res[3] -= num[4]*4, res[2] -= num[4]*4, res[1] -= num[4]*16;
if (res[3] < num[3]) {cout << "No\n"; return ;}
res[1] -= num[3]*9;
int mod = num[3] % 4, cnt = num[3]/4;
if (mod == 0) res[2] -= cnt * 9;
else if (mod == 3) res[2] -= cnt * 8;
else if (mod == 2) res[2] -= cnt * 6;
else if (mod == 1) res[2] -= cnt * 4;
if (res[2] < num[2]) {cout << "No\n"; return ;}
res[1] -= num[2]*4;
if (res[1] < num[1]) {cout << "No\n"; return ;}
cout << "Yes\n";
}
signed main () {
int t;
cin >> t;
while (t --) solve ();
}
