2025-11-15

Problem - 1858B - Codeforces（1500）（贪心）

这题一个是要读题
在不包括cookie seller的区间才算没吃饼干的时间sum += (s[i] - s[i - 1] - 1) / d;
在加上路过cookie seller时吃的饼干数 sum += m - 1;
然后贪心的思想，遍历删除每个cookie seller
重新计算合并两个区间里吃的饼干数

注意一点，要再处理s[0]=1-d和s[m+1]=n+1,这里就可以考虑到1~s[1]和s[m]~n区间

#include <bits/stdc++.h>
using namespace std;
#define LL long long
const LL mod = 998244353;
const int N=2e5+10;

void solve()
{
    int n, m, d;
    cin >> n >> m >> d;
    vector<int> s(m + 2);
    for (int i = 1; i<=m;i++){
        cin >> s[i];
    }
    s[0] = 1 - d;
    s[m + 1] = n + 1;
    int sum = 0;
    for (int i = 1; i <= m+1;i++){
        sum += (s[i] - s[i - 1] - 1) / d;
    }
    sum += m - 1;
    int ans = n + 1, cnt = 0;
    for (int i = 1; i <= m;i++){
        int res = sum;
        res -= (s[i] - s[i - 1] - 1) / d;
        res -= (s[i + 1] - s[i] - 1) / d;
        res += (s[i + 1] - s[i - 1] - 1) / d;
        if(res<ans){
            ans = res;
            cnt = 1;
        }else if(res==ans){
            cnt++;
        }
    }
    cout << ans << " " << cnt << endl;
}

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int T;
    cin >> T;
    while (T--)
    {
        solve();
    }
}

Problem - 1385D - Codeforces(string)

分治思想，递归

#include <bits/stdc++.h>
using namespace std;
#define LL long long
const LL mod = 998244353;
const int N=2e5+10;
string s;
int getans(int l, int r, char c)
{
    if (l == r)
        return s[l] != c;//相等返回0
    int tot1 = 0, tot2 = 0;
    int mid = (l + r) >> 1;
    for (int i = l; i <= mid; i++)
        if (s[i] != c)
            tot1++;
    for (int i = mid + 1; i <= r; i++)
        if (s[i] != c)
            tot2++;
    tot1 += getans(mid + 1, r, c + 1);
    tot2 += getans(l, mid, c + 1);
    return min(tot1, tot2);
}

void solve()
{
    int n;
    cin >> n >> s;
    cout << getans(0, n - 1, 'a') << endl;
}

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int T;
    cin >> T;
    while (T--)
    {
        solve();
    }
}

Problem - 873B - Codeforces（01string）（1500）

这题思路简单，存第一次出现数字下标，然后求前缀长度即可

#include <bits/stdc++.h>
using namespace std;
#define LL long long
const LL mod = 998244353;
const int N=2e5+10;
int sum[N], f[N], ans;

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int n;
    cin >> n;
    string s;
    cin >> s;
    s = " " + s;
    for (int i = 1; i <= n;i++){
        sum[i] = sum[i - 1] + (s[i] == '1' ? 1 : -1);
        if(sum[i]==0){
            ans = max(ans, i);
        }
    }
    for (int i = 1; i <= n;i++){
        if(f[sum[i]+n])
            ans = max(ans, i - f[sum[i]+n]);
        else
            f[sum[i]+n] = i;
    }
    cout << ans << endl;
}

昨天要做的洛谷题

P2347 [NOIP 1996 提高组] 砝码称重 - 洛谷

01背包问题

#include <bits/stdc++.h>
using namespace std;
#define LL long long
const LL mod = 998244353;
const int N=1010;
int a[N], x, num,ans;
int b[10] = {0, 1, 2, 3, 5, 10, 20};
bool vis[N];

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int x;
    for (int i = 1; i <= 6;i++){
        cin >> x;
        for (int j = 1; j <= x;j++)
            a[++num] = b[i];
    }
    vis[0] = 1;
    for (int i = 1; i <= num;i++){
        for (int j = 1000; j >= 0;j--)
            if(vis[j])
                vis[j + a[i]] = 1;
    }
    for (int i = 1; i <= 1000;i++){
        if(vis[i])
            ans++;
    }
    cout << "Total=" << ans << endl;
}

bitset解法
用二进制求方案数

#include <bits/stdc++.h>
using namespace std;
#define LL long long
const LL mod = 998244353;
const int N=2e5+10;
int a[10], w[10] = {1, 2, 3, 5, 10, 20};
bitset<1010> s;

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    for (int i = 0; i < 6;i++){
        cin >> a[i];
    }
    s[0] = 1;
    for (int i = 0; i < 6;i++){
        for (int j = 0; j < a[i];j++){
            s |= s << w[i];
        }
    }
    cout << "Total=" << s.count() - 1;
}