题目链接
https://www.patest.cn/contests/gplt/L1-032
思路
要分两种情况处理
①字符串长度 <= 填充长度
就在字符串前面输出(填充长度 - 字符串长度)个长度的填充字符
然后输出字符串
②字符串长度 > 填充长度
直接输出 (字符串长度 - 填充长度)个长度的后缀字符串字符就可以
AC代码
#include <cstdio>
#include <cstring>
#include <ctype.h>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#include <cmath>
#include <deque>
#include <vector>
#include <queue>
#include <string>
#include <map>
#include <stack>
#include <set>
#include <numeric>
#include <sstream>
using namespace std;
typedef long long LL;
const double PI = 3.14159265358979323846264338327;
const double E = 2.718281828459;
const double eps = 1e-6;
const int MAXN = 0x3f3f3f3f;
const int MINN = 0xc0c0c0c0;
const int maxn = 1e5 + 5;
const int MOD = 1e9 + 7;
int main()
{
string s;
int n;
char vis;
scanf("%d %c ", &n, &vis);
getline(cin, s);
int len = s.size();
if (len <= n)
{
n -= len;
while (n--)
cout << vis;
cout << s << endl;
}
else
{
for (int i = len - n; i < len; i++)
cout << s[i];
cout << endl;
}
}