PAT A1031 Hello World for U
A1031 Hello World for U
题目
如何求得n1、n2值
根据2*n1 + n2 = N + 2,对( N + 2 )/3 进行讨论：
- ( N + 2 )/3 = 0 ，n1 = n2 = ( N + 2 )/3
- ( N + 2 )/3 = 1，保证 n2 >= n1，多余的1分给n2
- ( N + 2 )/3 = 2，保证 n2 >= n1，多余的2分给n2
综上n1 = ( N + 2 ) / 3，n2 = ( N + 2 ) / 3 + ( N + 2 ) % 3
n1 = (length + 2) / 3;
n2 = (length + 2) / 3 + (length + 2) % 3;
参考代码
#include <cstdio>
#include <cstring>
int main() {
char s[10005];
scanf("%s", s);
int length = strlen(s);
int n1, n2, spacecounts;
n1 = (length + 2) / 3;
n2 = (length + 2) / 3 + (length + 2) % 3;
spacecounts = n2 - 2;
for (int i = 0; i < n1 - 1; ++i) {
printf("%c", s[i]);
for (int j = 0; j < spacecounts; ++j) {
printf(" ");
}
printf("%c\n", s[length - 1 - i]);
}
for (int i = 0; i < n2; ++i) {
printf("%c", s[n1 - 1 + i]);
}
return 0;
}
