Codeforces Round #713 (Div. 3)
Codeforces Round #713 (Div. 3)
倒霉掉分场, 刚出两个题, 电脑不干了...好在前两题手速快,没有掉太多分.
A题 Spy Detected!
题意
给一个序列, 序列中有一个数与其他数不同, 找到这个数的下标
思路
先在前三个数里面找到序列中相同的数是哪个, 然后遍历一遍就行了.
代码
#include <iostream>
#include <cstring>
#include <algorithm>
#include <map>
#include <set>
using namespace std;
int a[110];
void solve()
{
int n;
int ans;
cin >> n;
for (int i = 1; i <= n; i ++ )
cin >> a[i];
int t;
if (a[1] == a[2])
t = a[1];
else{
if (a[1] == a[3])
ans = 2;
else if (a[2] == a[3])
ans = 1;
cout << ans << endl;
return;
}
for (int i = 1; i <= n; i ++ ){
if (a[i] != t)
ans = i;
}
cout << ans << endl;
return;
}
int main()
{
int t;
cin >> t;
while (t -- )
solve();
return 0;
}
B题 Almost Rectangle
题意
给定一个方阵, 再给定两个点, 要求再寻找两个点, 是这四个点构成一个矩形.
思路
直接找就行了, 注意当已知的两个点为同行或同列时, 注意判断是否出界
代码
#include <iostream>
#include <cstring>
#include <algorithm>
#include <map>
#include <set>
using namespace std;
const int N = 410;
char s[N][N];
int n;
void print()
{
for (int i = 1; i <= n; i ++ ){
for (int j = 1; j <= n; j ++ )
cout << s[i][j];
cout << endl;
}
}
void solve()
{
int a = 0;
int b = 0;
int c = 0;
int d = 0;
cin >> n;
for (int i = 1; i <= n; i ++ )
cin >> s[i] + 1;
for (int i = 1; i <= n; i ++ )
for (int j = 1; j <= n; j ++ ){
if (a == 0 && s[i][j] == '*'){
a = i;
b = j;
}else if (c == 0 && s[i][j] == '*'){
c = i;
d = j;
}
}
if (a == c){
if (a > 1){
s[a - 1][b] = '*';
s[c - 1][d] = '*';
}else{
s[a + 1][b] = '*';
s[c + 1][d] = '*';
}
print();
return;
}
if (b == d){
if (b > 1){
s[a][b - 1] = '*';
s[c][d - 1] = '*';
}else{
s[a][b + 1] = '*';
s[c][d + 1] = '*';
}
print();
return;
}
s[a][d] = '*';
s[c][b] = '*';
print();
return;
}
int main()
{
int t;
cin >> t;
while (t -- )
solve();
return 0;
}
C题 A-B Palindrome
题意
给定一个包含\(01?\)的序列, 要求将\(?\)换为\(1或0\),使得该序列为回文序列, 且一定有\(a\)个0, \(b\)个1.
思路
遍历两边:
第一遍:判断是否有已经不合法的位置(\(1->0\)), 同时将可以确定的\(?\)确定.
第二遍:将剩余位置的\(?\)转化为\(0或1\), 同时判断是否出现不合法情况
代码
#include <iostream>
#include <cstring>
#include <algorithm>
#include <string>
using namespace std;
const int N = 2e5 + 10;
char s[N];
int a, b;
int n;
int mid;
void check()
{
for (int i = 1; i < mid; i ++ ){
if (s[i] == s[n - i + 1]){
if (s[i] == '1')
b -= 2;
if (s[i] == '0')
a -= 2;
}else{
if (s[i] != '?' && s[n - i + 1] != '?'){
puts("-1");
return;
}else{
if (s[i] == '?'){
s[i] = s[n - i + 1];
if (s[i] == '1')
b -= 2;
else if (s[i] == '0')
a -= 2;
}
if (s[n - i + 1] == '?'){
s[n - i + 1] = s[i];
if (s[i] == '1')
b -= 2;
else if (s[i] == '0')
a -= 2;
}
}
}
}
for (int i = 1; i < mid; i ++ ){
if (s[i] == s[n - i + 1]){
if (s[i] == '?'){
if (a && a != 1){
s[i] = s[n - i + 1] = '0';
a -= 2;
}else if (b && b != 1){
s[i] = s[n - i + 1] = '1';
b -= 2;
}
}
}
}
if (n % 2){
if (s[mid] == '1'){
b--;
}
if (s[mid] == '0')
a--;
if (s[mid] == '?'){
if (a){
s[mid] = '0';
a--;
}
if (b){
s[mid] = '1';
b--;
}
}
}
int flag = 1;
for (int i = 1; i <= n; i ++ )
if (s[i] == '?'){
flag = 0;
break;
}
if (a != 0 || b != 0 || flag == 0)
puts("-1");
else{
cout << s + 1 << endl;
}
}
void solve()
{
cin >> a >> b;
scanf("%s", s + 1);
n = a + b;
if (n % 2)
mid = (1 + n) / 2;
else
mid = (1 + n) / 2 + 1;
/*
for (int i = 1; i < mid; i ++ ){
if (s[i] == s[n - i + 1]){
if (s[i] == '1')
b -= 2;
if (s[i] == '0')
a -= 2;
}else{
if (s[i] != '?' && s[n - i + 1] != '?'){
puts("-1");
return;
}else{
if (s[i] == '?'){
s[i] = s[n - i + 1];
if (s[i] == '1')
b -= 2;
else if (s[i] == '0')
a -= 2;
}
if (s[n - i + 1] == '?'){
s[n - i + 1] = s[i];
if (s[i] == '1')
b -= 2;
else if (s[i] == '0')
a -= 2;
}
}
}
}
if (n % 2)
if (s[mid] != '?'){
if (s[mid] == '0')
a--;
else
b--;
}
*/
check();
//cout << s + 1 << "-->" << a << ' ' << b << endl;
return;
}
int main()
{
int t;
cin >> t;
while (t -- )
solve();
return 0;
}
D题 Corrupted Array
题意
给定一个长度为\(n + 2\)的序列, 在序列中找到一个数\(k\)为剩余\(n + 1\)个数中的\(n\)个数之和.
思路
贪心, 首先对序列排序, 可以确定\(k\)一定为序列最大值或次大值:
证明:
- 若\(k\)不是最大值或次大值, 那么剩余的\(n + 1\)个数, 无论怎么选\(n\)个数, 一定会产生和大于\(k\).即\(k\)必须为最大或次大值.
- \(k\)为次大值, 则\(x\)必须为最大值, 理由同上, 这时只需判断剩余数字之和是否为\(k\)
- \(k\)为最大值, 先求出剩余\(n + 1\)个数字之和, 逐个判断\(x\),是否存在\(x\).
#include <iostream>
#include <cstring>
#include <algorithm>
#include <map>
#include <set>
using namespace std;
const int N = 2e5 + 10;
int b[N];
int a[N];
void solve()
{
int n;
cin >> n;
for (int i = 1; i <= n + 2; i ++ )
scanf("%d", &b[i]);
sort(b + 1, b + n + 3);
long long k = b[n + 2];
long long y = b[n + 1];
long long sum = 0;
for (int i = 1; i <= n + 1; i ++ )
sum += b[i];
//cout << sum << ' ' << k << ' ' << y << endl;
//选次大值:
if (sum - y == y){
for (int i = 1; i <= n; i ++ )
printf("%d ", b[i]);
puts("");
return;
}
//选最大值:
int del = 0;
for (int i = 1; i <= n + 1; i ++ )
if (sum - b[i] == k){
del = i;
break;
}
if (del != 0){
for (int i = 1; i <= n + 1; i ++ )
if (i != del)
printf("%d ", b[i]);
puts("");
return;
}
puts("-1");
return;
}
int main()
{
int t;
cin >> t;
while (t -- )
solve();
return 0;
}
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