26.P1628 合并序列
法一：
#include<bits/stdc++.h>
using namespace std;
const int N = 1e5 + 10;
int n, cnt;
string c, s[N], a[N];
int main()
{cin >> n;for(int i = 1; i <= n; i ++){cin >> s[i];}cin >> c;int len = c.size();for(int i = 1; i <= n; i ++){if(s[i].substr(0, len) == c){cnt ++;a[cnt] = s[i];}}sort(a + 1, a + 1 + cnt);for(int i = 1; i <= cnt; i ++){cout << a[i] << '\n';}return 0;
}法二：
使用find函数，其是指向搜索范围内第一个元素
#include<bits/stdc++.h>
using namespace std;
const int N = 1e5 + 10;
int n, cnt;
string c, s[N], a[N];
int main()
{cin >> n;for(int i = 1; i <= n; i ++){cin >> s[i];}cin >> c;sort(s + 1, s + 1 + n);for(int i = 1; i <= n; i ++){if(s[i].find(c) == 0)cout << s[i] << '\n';}return 0;
}27.P1403 [AHOI2005] 约数研究
约数个数 = 将约数拆解成每个质因子乘积的形式，每次将质因子出现的个数+1后相乘得到约数个数，普通一次一次模板计算会有超时现象
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 110, mod = 1e9 + 7;
ll n, ans, primes[N];
ll solve(int x)
{unordered_map<int, int> primes;for(int i = 2; i <= x / i; i ++){while(x % i == 0){x /= i;primes[i] ++;}}if(x > 1)primes[x] ++;ll res = 1;for(auto i : primes){res = res * (i.second + 1) % mod;}return res;
}
int main()
{cin >> n;for(int i = 1; i <= n; i ++){ans += solve(i);}cout << ans;return 0;
}AC写法：
每次到自己就将自己可以贡献的所有约数个数不断加起来
#include<bits/stdc++.h>
using namespace std;
int n, ans;
int main()
{cin >> n;for(int i = 1; i <= n; i ++){for(int j = i; j <= n; j += i){ans ++;}}cout << ans;return 0;
}28.P1644 跳马问题
法一：
#include<bits/stdc++.h>
using namespace std;
int dx[4] = {-2, -1, 1, 2};
int dy[4] = {1, 2, 2, 1};
int n, m, ans;
void dfs(int x, int y)
{if(x == m && y == n){ans ++;return;}for(int i = 0; i < 4; i ++){int a = x + dx[i];int b = y + dy[i];if(a >= 0 && a <= m && b >= 0 && b <= n){dfs(a, b);}}
}
int main()
{cin >> m >> n;dfs(0, 0);cout << ans;return 0;
}法二：
#include<bits/stdc++.h>
using namespace std;
int dy[4] = {-2, -1, 1, 2};
int dx[4] = {- 1, - 2, - 2, - 1};
const int N = 1e3 + 10;
int n, m, dp[N][N];
int main()
{cin >> n >> m;dp[0][0] = 1;for(int i = 0; i <= m; i ++){for(int j = 0; j <= n; j ++){for(int k = 0; k < 4; k ++){int a = i + dx[k];int b = j + dy[k];if(a < 0 || a > m || b < 0 || b > n)continue;dp[i][j] += dp[a][b];}}}cout << dp[m][n];
}29.P1122 最大子树和
将每朵花的两条边相连，进行dfs,传入两个参数即为当前节点与当前节点的父亲节点，dp[x]为当前节点的值，将与当前节点的值循环遍历一遍，如果现节点是当前节点的父亲则跳过，因为其是不断向下递归，只用向下的点是大于0的则一定会对答案有所贡献，现在每一个dp的值都代表了自己所能到达的最大范围，最后将每一个点遍历，找出最大的那个可以取到的和
#include<bits/stdc++.h>
using namespace std;
const int N = 1e6 + 10;
int n, u, v, w[N], dp[N];
vector<int> g[N];
void dfs(int x, int fa)
{dp[x] = w[x];for(auto y : g[x]){if(y == fa)continue;dfs(y, x);if(dp[y] > 0)dp[x] += dp[y];}
}
int main()
{cin >> n;for(int i = 1; i <= n; i ++)cin >> w[i];for(int i = 1; i <= n - 1; i ++){cin >> u >> v;g[u].push_back(v);g[v].push_back(u);}dfs(1, 0);int ans = w[1];for(int i = 1; i <= n; i ++){ans = max(ans, dp[i]);}cout << ans;return 0;
}30.P1510 精卫填海
dp[i]表示到目前为止，得到总体积为i的木石所需的最小体力，输出最大体力相减即可
#include<bits/stdc++.h>
using namespace std;
const int N = 1e5 + 10;
int v, n, c, k[N], m[N], dp[N];
//dp[i]表示到目前为止，得到总体积为i的木石所需的最小体力
int main()
{cin >> v >> n >> c;for(int i = 1; i <= n; i ++){cin >> k[i] >> m[i];}memset(dp, 0x3f, sizeof dp);dp[0] = 0;for(int i = 1; i <= n; i ++){for(int j = 1e4 + 10; j >= k[i]; j --){dp[j] = min(dp[j], dp[j - k[i]] + m[i]);}}int ans = dp[v];for(int i = v; i <= 2e4; i ++){ans = min(ans, dp[i]);}if(c >= ans)cout << c - ans;else cout << "Impossible" << '\n';return 0;
}
