A. Line Breaks
思路:从头开始累加单词个数,超过m就退出。
代码:
#include <bits/stdc++.h>
using namespace std;
#define int long long
void solve()
{
int n, m, k;
cin >> n >> m;
vector<string> a(n);
int cnt = 0;
for (int i = 0; i<n; i++){
cin >> a[i];
}
int ans = 0;
for (int i = 0; i<n; i++){
if (cnt+a[i].size() <= m){
cnt += a[i].size();
ans++;
}
else break;
}
cout << ans << endl;
}
signed main()
{
ios::sync_with_stdio(false);
cin.tie(0); cout.tie(0);
int t = 1;
cin >> t;
while (t--)
{
solve();
}
return 0;
}
B. Transfusion
思路:分析得出,应该对奇数和偶数分开讨论。
代码:
#include <bits/stdc++.h>
using namespace std;
#define int long long
#define vi vector<int>
#define pb push_back
void solve()
{
int n, m, k;
cin >> n;
int sum1= 0, sum2=0;
vi a, b;
for (int i = 1;i<=n; i++){
int x; cin >> x;
if (i % 2 == 1){
a.pb(x);
sum1 += x;
}
else{
b.pb(x);
sum2 += x;
}
}
if (sum1%a.size() == 0 && sum2%b.size() == 0 && sum1/a.size() == sum2/b.size()){
cout << "YES" << endl;
}
else{
cout << "NO" << endl;
}
}
signed main()
{
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
int t = 1;
cin >> t;
while (t--)
{
solve();
}
return 0;
}
C. Uninteresting Number
思路:回想起了初中还是小学知识,一个是9的倍数的数,各个位数上的和也是9的倍数。
在这里我们遍历累加n的位数,又由题可以推出我们只可能当遇到2,3的时候才能用它的平方数替代,但是我们不知道该替代哪些,所以先分别记录下2,3的个数。
分别嵌套循环遍历2,3从0取到完为止的数是否满足要求。
代码:
#include <bits/stdc++.h>
using namespace std;
#define int long long
void solve()
{
int n, m, k;
string s;
cin >> s;
int a = 0, b = 0, sum = 0;
for (char c : s)
{
if (c == '2') a++;
else if (c == '3') b++;
sum += c - '0';
}
for (int i = 0; i <= a; i++){
for (int j = 0; j <= b; j++){
if ((sum + i * 2 + j * 6) % 9 == 0){
cout << "YES" << endl;
return;
}
}
}
cout << "NO" << endl;
}
signed main()
{
ios::sync_with_stdio(false);
cin.tie(0); cout.tie(0);
int t = 1;
cin >> t;
while (t--)
{
solve();
}
return 0;
}
D. Digital string maximization
思路:感觉就像是冒泡排序一样的思路,不过改一点点东西而已。(一开始完全没想到这样会过,想半天才试了以下居然能过)。在排序的思路上对要往上的-1进行比较和替换。看代码比较好理解。
代码:
#include <bits/stdc++.h>
using namespace std;
#define int long long
void solve()
{
int n, m, k;
string s;
cin >> s;
int num = s.size();
for (int i = 0; i < num; i++){
for (int j = i; j>=1; j--){
if (s[j]-1 > s[j-1]){
char temp = s[j]-1;
s[j] = s[j-1];
s[j-1] = temp;
}
else break;
}
}
cout << s << endl;
}
signed main()
{
ios::sync_with_stdio(false);
cin.tie(0); cout.tie(0);
int t = 1;
cin >> t;
while (t--)
{
solve();
}
return 0;
}
E. Three Strings
思路:一道dp题,思路很像编辑距离和最长公共子序列,可以去刷刷这两题。
代码:
#include <bits/stdc++.h>
using namespace std;
#define int long long
#define vi vector<int>
#define pb push_back
#define endl '\n'
void solve() {
string a, b, c;
cin >> a >> b >> c;
int n = a.size(), m = b.size(), k = c.size();
vector<vector<int>> dp(n + 1, vector<int>(m + 1, LLONG_MAX));
dp[0][0] = 0;
for (int i = 0; i <= n; ++i) {
for (int j = 0; j <= m; ++j) {
if (i + j <= k) {
if (i > 0) {
dp[i][j] = min(dp[i][j], dp[i - 1][j] + (c[i + j - 1] != a[i - 1]));
}
if (j > 0) {
dp[i][j] = min(dp[i][j], dp[i][j - 1] + (c[i + j - 1] != b[j - 1]));
}
}
}
}
int result = LLONG_MAX;
for (int i = 0; i <= n; ++i) {
if (i <= k) {
result = min(result, dp[i][m] + (k - (i + m)));
}
}
cout << result << endl;
}
signed main() {
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
int t;
cin >> t;
while (t--) {
solve();
}
return 0;
}
F. Maximum modulo equality
思路:看到求区间类的题应该要很快想到线段树,这题应该最大公因数满足结合律,也可以用ST表来做比较快。
代码:
#include <bits/stdc++.h>
using namespace std;
#define int long long
#define vi vector<int>
void solve()
{
int n, m, k, q;
cin >> n >> q;
vi a(n);
vector<vector<int>> b(n, vi(20));
for (int i = 0; i<n; i++) {
cin >> a[i];
if (i) b[i][0] = abs(a[i]-a[i-1]);
}
for (int j = 1; j<20; j++){
for (int i = 0; i+(1ll<<j)-1<n; i++){
b[i][j] = __gcd(b[i][j-1], b[i+(1ll<<j-1)][j-1]);
}
}
auto query = [&](int l, int r)->int
{
if (l > r) return 0;
int k = log2(r-l+1);
return __gcd(b[l][k], b[r-(1ll<<k)+1][k]);
};
while (q--){
int l, r;
cin >> l >> r;
l--, r--;
cout << abs(query(l+1, r)) << " ";
}
cout << endl;
}
signed main()
{
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
int t = 1;
cin >> t;
while (t--)
{
solve();
}
return 0;
}
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