Better_Tomorrow @ 2024-05-19 19:31:22
#include<bits/stdc++.h>
using namespace std;
int main() {
long long k;
int sum = 0;
string a;
cin >> a >> k;
for (int i = 0; i < a.size(); i++) {
if (a[i] - '0' == 3) {
sum++;
}
}
if (sum == k) {
cout << "yes";
} else {
cout << "no";
}
return 0;
}
by Better_Tomorrow @ 2024-05-19 19:33:06
我看题解全用的是while,可我跟题解的思路应该没错吧?
by QBW1117 @ 2024-05-19 19:33:23
输出 YES,否则,输出 NO。
by QBW1117 @ 2024-05-19 19:33:32
大写
by ZhangXuKun @ 2024-05-19 19:45:40
如果输入的k大于字符串a的长度,那么即使字符串a中包含3的数量等于k,程序也会输出"no"
by Lizichen_licis @ 2024-05-19 19:47:17
@QBW1117