A
題意: 假設(shè)字符串中出現(xiàn)次數(shù)最少的字母是x, 出現(xiàn)次數(shù)為y, 刪除所有出現(xiàn)次數(shù)為y的字符
思路: 統(tǒng)計(jì)一下字符的出現(xiàn)次數(shù), 然后照著做就行

#include <iostream> #include <algorithm> #include <string> #include <cmath> #include <cstring> #include <stack> #include <queue> #include <utility> #include <map> #include <set> #include <cstdio> #include <sstream> #include <cctype>#define LL long long #define mst(a, b) memset(a, b, sizeof(a)) #define pill pait<int, int> #define ft first #define sd secondusing namespace std;const int qq = 3e5 + 10; const int MOD = 1e9 + 7; const int INF = 1e9 + 10; int vis[300]; char s[qq];int main() {string x; cin >> x;int minx = INF;for (int i = 0; i < x.size(); ++i) {vis[x[i]]++;}string ans = "";for (int i = 0; i < x.size(); ++i) {if (vis[x[i]] > 0) minx = min(minx, vis[x[i]]);}for (int i = 0; i < x.size(); ++i) {if (minx == vis[x[i]]) continue;ans += x[i];}cout << ans << endl;return 0; }

B
題意: 求第n個(gè)丑數(shù)
思路: leetcode原題, 可以去leetcode找一找

#include <iostream> #include <algorithm> #include <string> #include <cmath> #include <cstring> #include <stack> #include <queue> #include <utility> #include <map> #include <set> #include <cstdio> #include <sstream> #include <cctype>#define LL long long #define mst(a, b) memset(a, b, sizeof(a)) #define pill pait<int, int> #define ft first #define sd secondusing namespace std;const int qq = 3e5 + 10; const int MOD = 1e9 + 7; const int INF = 1e9 + 10; int n; int num[qq]; void init() {num[0] = 1;int a, b, c;a = b = c = 0;int top = 0;for (int i = 1; top <= 1500; ++i) {int x = num[a] * 2;int y = num[b] * 3;int z = num[c] * 5;if (x <= y && x <= z) {num[++top] = x, a++;} else if (y <= x && y <= z) {num[++top] = y, b++;} else {num[++top] = z, c++;}while (num[a] * 2 <= num[top]) a++;while (num[b] * 3 <= num[top]) b++;while (num[c] * 5 <= num[top]) c++;}// for (int i = 0; i < 9; ++i) {// printf("%d ", num[i]);// }// puts(""); }int main() {init();scanf("%d", &n);printf("%d\n", num[n - 1]);return 0; }

C
題意: 求兩個(gè)串前后重疊部分的最大長度
思路: 最初我是采用這種暴力計(jì)算的方法, 可以ac

#include <iostream> #include <algorithm> #include <string> #include <cmath> #include <cstring> #include <stack> #include <queue> #include <utility> #include <map> #include <set> #include <cstdio> #include <sstream> #include <cctype>#define LL long long #define mst(a, b) memset(a, b, sizeof(a)) #define pill pait<int, int> #define ft first #define sd secondusing namespace std;const int qq = 3e5 + 10; const int MOD = 1e9 + 7; const int INF = 1e9 + 10; string x, y;int main() {std::ios::sync_with_stdio(false);cin >> x >> y;int maxn = 0;for (int i = x.size() - 1; i >= 0; --i) {int start = i, len = 0;for (int j = 0; j < y.size() && start < x.size(); ++j) {if (x[start] == y[j]) {start++, len++;} else {break;}}maxn = max(maxn, len);}for (int i = y.size() - 1; i >= 0; --i) {int start = i, len = 0;for (int j = 0; j < x.size() && start < y.size(); ++j) {if (y[start] == x[j]) {start++, len++;} else {break;}}maxn = max(maxn, len);}cout << maxn << endl;return 0; }

之后想了想發(fā)現(xiàn)其實(shí)就是個(gè)kmp問題, 跑兩次kmp即可 時(shí)間復(fù)雜度O(n)

#include <iostream> #include <algorithm> #include <string> #include <cmath> #include <cstring> #include <stack> #include <queue> #include <utility> #include <map> #include <set> #include <cstdio> #include <sstream> #include <cctype>#define LL long long #define mst(a, b) memset(a, b, sizeof(a)) #define pill pait<int, int> #define ft first #define sd secondusing namespace std;const int qq = 3e5 + 10; const int MOD = 1e9 + 7; const int INF = 1e9 + 10; string x, y; int nt[qq]; void getNext(string f) {int i = 0, j = -1;nt[0] = -1;int len = f.size();while (i < len) {if (j == -1 || f[i] == f[j]) {nt[++i] = ++j;} else {j = nt[j];}} } int getMaxn(string x, string y) {int i = 0, j = 0;int len = x.size();while (i < len) {if (j == -1 || x[i] == y[j]) {i++, j++;} else {j = nt[j];}}return j; }int main() {std::ios::sync_with_stdio(false);cin >> x >> y;getNext(x);int maxn = getMaxn(y, x);getNext(y);maxn = max(maxn, getMaxn(x, y));cout << maxn << endl;return 0; }

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