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这个题我感觉整体难度并不算大,但是就是有一点麻烦,因为要考虑三种情况,给他增加一位还是两位还是三位,因为495为三位数,所以说增加最多增加三位就肯定可以满足情况了。并且我们要考虑一种情况,就是当这个数就是495的倍数时,就直接输出-1就可以了。
下面我们来看一下代码:
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
int main() {
ios::sync_with_stdio(0);
cin.tie(0); cout.tie(0);
ll n; cin >> n;
if (n % 495 == 0) return cout << -1 << endl, 0;
for (int i = 0; i <= 9; i ++ )
if ((10 * n + i) % 495 == 0) {
cout << i << endl;
return 0;
}
for (int i = 0; i <= 9; i ++ )
for (int j = 0; j <= 9; j ++ ) {
if ((100 * n + 10 * i + j) % 495 == 0) {
cout << i << j << endl;
return 0;
}
}
for (int i = 0; i <= 9; i ++ )
for (int j = 0; j <= 9; j ++ )
for (int k = 0; k <= 9; k ++ ) {
if ((1000 * n + 100 * i + 10 * j + k) % 495 == 0) {
cout << i << j << k << endl;
return 0;
}
}
return 0;
}
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
int main() {
ios::sync_with_stdio(0);
cin.tie(0); cout.tie(0);
ll n; cin >> n;
if (n % 495 == 0) return cout << -1 << endl, 0;
for (int i = 0; i <= 9; i ++ )
if ((10 * n + i) % 495 == 0) {
cout << i << endl;
return 0;
}
for (int i = 0; i <= 9; i ++ )
for (int j = 0; j <= 9; j ++ ) {
if ((100 * n + 10 * i + j) % 495 == 0) {
cout << i << j << endl;
return 0;
}
}
for (int i = 0; i <= 9; i ++ )
for (int j = 0; j <= 9; j ++ )
for (int k = 0; k <= 9; k ++ ) {
if ((1000 * n + 100 * i + 10 * j + k) % 495 == 0) {
cout << i << j << k << endl;
return 0;
}
}
return 0;
}
#include <stdio.h>
int main() {
long long n;
scanf("%lld", &n);
if (n % 495 == 0) {
printf("-1\n");
return 0;
}
for (int i = 0; i <= 9; i++) {
if ((10 * n + i) % 495 == 0) {
printf("%d\n", i);
return 0;
}
}
for (int i = 0; i <= 9; i++) {
for (int j = 0; j <= 9; j++) {
if ((100 * n + 10 * i + j) % 495 == 0) {
printf("%d%d\n", i, j);
return 0;
}
}
}
for (int i = 0; i <= 9; i++) {
for (int j = 0; j <= 9; j++) {
for (int k = 0; k <= 9; k++) {
if ((1000 * n + 100 * i + 10 * j + k) % 495 == 0) {
printf("%d%d%d\n", i, j, k);
return 0;
}
}
}
}
return 0;
}#include <stdio.h>
int main() {
long long n;
scanf("%lld", &n);
if (n % 495 == 0) {
printf("-1\n");
return 0;
}
for (int i = 0; i <= 9; i++) {
if ((10 * n + i) % 495 == 0) {
printf("%d\n", i);
return 0;
}
}
for (int i = 0; i <= 9; i++) {
for (int j = 0; j <= 9; j++) {
if ((100 * n + 10 * i + j) % 495 == 0) {
printf("%d%d\n", i, j);
return 0;
}
}
}
for (int i = 0; i <= 9; i++) {
for (int j = 0; j <= 9; j++) {
for (int k = 0; k <= 9; k++) {
if ((1000 * n + 100 * i + 10 * j + k) % 495 == 0) {
printf("%d%d%d\n", i, j, k);
return 0;
}
}
}
}
return 0;
}