9.10 中国电信it研发中心 笔试编程题
A
题意: 假设字符串中出现次数最少的字母是x, 出现次数为y, 删除所有出现次数为y的字符
思路: 统计一下字符的出现次数, 然后照着做就行
#include
#define ft first
#define sd second
using 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
题意: 求第n个丑数
思路: leetcode原题, 可以去leetcode找一找
#include
#define ft first
#define sd second
using 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
题意: 求两个串前后重叠部分的最大长度
思路: 最初我是采用这种暴力计算的方法, 可以ac
#include
#define ft first
#define sd second
using 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;
}之后想了想发现其实就是个kmp问题, 跑两次kmp即可 时间复杂度O(n)
#include
#define ft first
#define sd second
using 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;
}