去打cometoj的比赛去了qwq然后这场开局就有点晚...
估计小号涨不了啥分。
atc啥时候才有arc啊....
A
#include
using namespace std;
int main() {
int a, b;
cin >> a >> b;
int ans = max(a * 2 - 1, max(b * 2 - 1, a + b));
cout << ans;
} B
#include
using namespace std;
int f[110], a[110], ans = 0;
int main() {
int n; cin >> n;
for(int i = 1; i <= n; ++i) cin >> a[i];
int now = 0;
for(int i = 1; i <= n; ++i) {
if(now <= a[i]) ++ans;
now = max(now, a[i]);
}
cout << ans;
} C
合法排列情况就两种...分别模拟取min即可。
#include
using namespace std;
#define ll long long
const int N = 1000100;
char s[N];
int main() {
scanf("%s", s+1);
int now = 0, n = strlen(s+1), sum = 0, ans = 0;
for(int i = 1; i <= n; ++i) {
now ^= 1;
if(s[i] - '0' != now) ++ans;
}
now = 1;
for(int i = 1; i <= n; ++i) {
now ^= 1;
if(s[i] - '0' != now) ++sum;
}
printf("%d\n", min(ans, sum));
} D
一开始看错题意写了个堆求1的最多次数的。然后发现是连续的，这个玩意是有单调性的所以直接双指针/二分即可。注意要先缩点，把相同的缩成一块不然不好算。
#include
using namespace std;
const int N = 100010;
char s[N];
int a[N], op[N], tot[N];
int main() {
int n, k, cnt = 0, sum = 0; cin >> n >> k;
scanf("%s", s+1);
for(int i = 1; i <= n + 1; ++i) {
if(s[i] != s[i - 1] && i - 1) {
a[++cnt] = sum;
op[cnt] = s[i - 1] - '0';
sum = 0;
}
++sum;
}
memset(tot, 0, sizeof(tot));
for(int i = 1; i <= cnt; ++i) {
if(!op[i]) tot[i] = tot[i - 1] + 1; else tot[i] = tot[i - 1];
a[i] += a[i - 1];
}
int ans = 0;
for(int l = 0, r = 1; r <= cnt; ++r) {
while(l < r && tot[r] - tot[l - 1] > k) ++l;
ans = max(ans, a[r] - a[l - 1]);
}
printf("%d\n", ans);
}