Codeforces Round #700 (Div. 2) A ~ D1个人题解
Codeforces Round #700 (Div. 2)
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1480A. Yet Another String Game
因为Alice是要追求小,Bob追求大值,所以只要变为 a
或 z
.
AC代码
int main() {
ios_base::sync_with_stdio(false), cin.tie(0);
for (cin >> _; _--;) {
string s;
cin >> s;
for (int i = 0; i < s.size(); ++i)
if (i % 2 == 0) {
if (s[i] == 'a') cout << 'b';
else cout << 'a';
} else {
if (s[i] != 'z') cout << 'z';
else cout << 'y';
}
cout << "\n";
}
return 0;
}
1480B. The Great Hero
排序,对英雄伤害少的排前面,统计总共打败的敌人数即可
AC代码
using ll = long long;
int _;
struct node {
ll a, b;
};
ll A, B;
bool cmp(node a, node b) {
ll d1 = (a.b / A + a.b % A != 0) * a.a, d2 = (b.b / A + b.b % A != 0) * b.a;
return d1 < d2;
}
int main() {
ios_base::sync_with_stdio(false), cin.tie(0);
for (cin >> _; _--;) {
int n, cnt = 0;
cin >> A >> B >> n;
vector<node> e(n);
for (int i = 0; i < n; ++i) cin >> e[i].a;
for (int i = 0; i < n; ++i) cin >> e[i].b;
sort(e.begin(), e.end(), cmp);
for (int i = 0; i < n; ++i) {
while (B >= 1 && e[i].b >= 1) e[i].b -= A, B -= e[i].a;
if (e[i].b <= 0) cnt++;
}
cout << (cnt == n ? "YES\n" : "NO\n");
}
return 0;
}
1480C. Searching Local Minimum
C题没做出来,待补。
1480D1. Painting the Array I
题意和思路来自网络
个人理解以后的AC代码
int _;
int a[210000];
int main() {
ios_base::sync_with_stdio(false), cin.tie(0);
int n;
cin >> n;
for (int i = 1; i <= n; ++i) cin >> a[i];
int lst = 0, sum = 0;
for (int i = 1; i <= n; ++i) {
if (a[i] == a[i + 1]) {
if (a[lst] != a[i])
sum += 2;
else {
bool f = false;
for (int j = i - 1; true; --j) {
if (a[j] != a[i] && a[j + 1] != a[i]) {
f = true;
break;
}
if (j == lst) break;
}
sum += f ? 2 : 1;
}
while (a[i] == a[i + 1]) ++i;
lst = i;
} else
sum++;
// cout << sum << "\n";
}
cout << sum << "\n";
return 0;
}
1480D2. Painting the Array II
D2昨天没啥想法就被弃掉了(老菜鸡了)
顺便贴一下朋友的D2代码
AC代码
int _;
int a[100005], b[100005], ans[100005], now[100005], pre[100005];
int dfs(int x) {
if (ans[x] != -1) return ans[x];
if (x == 0) return 0;
ans[x] = dfs(x - 1);
if (pre[x]) ans[x] = max(ans[x], dfs(pre[x] + 1) + 1);
return ans[x];
}
int main() {
int T = 1;
// cin>>T;
while (T--) {
int n, x = 0;
cin >> n;
for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
for (int i = 1; i <= n; i++)
if (a[i] != a[i - 1]) b[++x] = a[i];
for (int i = 1; i <= x; i++) pre[i] = now[b[i]], now[b[i]] = i;
memset(ans, -1, sizeof(ans));
dfs(x);
int mx = 0;
for (int i = 1; i <= x; i++) mx = max(mx, ans[i]);
cout << x - mx << endl;
}
return 0;
}