文章目录
- 1.枚举
- 2.模拟
1.枚举
[例1] 特别数的和
评测系统
#include
using namespace std;
bool pa(int x) {
while (x) {
if (x % 10 == 2 || x % 10 == 1 || x % 10 == 0 || x % 10 == 9) {
return true;
}
else {
x = x / 10;
}
}
return false;
}
int main()
{
int sum=0;
int m;
cin >> m;
for (int i = 1; i m; i++) {
if (pa(i)) {
sum += i;
}
}
cout sum;
return 0;
}
[例2] 反序数
评测系统
#include
using namespace std;
int main()
{
int n;
cin >> n;
int a, b, c;
cin >> a >> b >> c;
int sum = 0;
for (int i = 1; i n; i++) {
if (i % a != 0 && i % b != 0 && i % c != 0) {
sum++;
}
}
cout sum;
return 0;
}
[例3] 找到最多的数
评测系统
#include
#include
#include
using namespace std;
int main()
{
int n, m;
cin >> n >> m;
int num = n * m;
mapint,int> mp;
vectorint> a(num);//记录去重后的序列
while (num--) {
int x;
cin >> x;
if (!mp.count(x)) {
a.push_back(x);
}
mp[x]++;
}
for (const auto& x:a) { //也可以不使用vector，每次从map中取出[x.y]
if (mp[x] * 2 > n * m)
cout x;
}
}
[例4]
评测系统
对于每种可能的颜色，我们尝试计算如果将走廊涂成这种颜色需要多少天。在遍历过程中要跳过已经是这种颜色的房子，每次移动k步。最后记录对于这种颜色所需的涂色天数，并与之前的最小天数进行比较，取较小值。
输入数据的最大值即为初始颜色种类数
#include
#include
using namespace std;
int main()
{
int t;
cin >> t;
while (t--) {
int n, k;
cin >> n >> k;
vectorint> a(n);
int maxnum = 0;//记录初始颜色种类数
for (int i = 0; i n; ++i) {
cin >> a[i];
maxnum=max(maxnum, a[i]);
}
int ans = n;
for (int c = 1; c maxnum; ++c) {
int cnt = 0;
int idx = 0;
while (idx n) {
if (a[idx] == c) {
idx++;
continue;
}
idx += k;
cnt++;
}
ans = min(cnt, ans);
}
cout ans endl;
}
}
2.模拟
[例1] 扫雷
评测系统
#include
using namespace std;
const int N = 150;
int main()
{
int n, m;
cin >> n >> m;
int a[N][N], ans[N][N];
for (int i = 0; i n; i++) {
for (int j = 0; j m; j++) {
cin >> a[i][j];
}
}
for (int i = 0; i n; i++) {
for (int j = 0; j m; j++) {
if (a[i][j] == 1) {
ans[i][j] = 9;
continue;
}
int i2 = i, j2 = j;//遍历四周
for (i2 = max(0, i - 1); i2 min(n - 1, i + 1); i2++) {
for (j2 = max(0, j - 1); j2 min(m - 1, j + 1); j2++) {
if (a[i2][j2] == 1) ans[i][j]++;
}
}
}
}
for (int i = 0; i n; i++) {
for (int j = 0; j m; j++) {
cout ans[i][j]" ";
}
cout endl;
}
return 0;
}
[例2] 灌溉
评测系统
使用类似于BFS的过程，用bool类型的irrigated数组记录当前位置是否已被灌溉，最终统计已被灌溉的数量。
#include
#include
using namespace std;
const int N = 200;
int main()
{
int n, m, t;
cin >> n >> m >> t;
bool irrigated[N][N];
queuepairint, int>> q;
for (int i = 0; i t; i++) {
int r, c;
cin >> r >> c;
r--;
c--;
irrigated[r][c] = true;
q.push({ r,c });
}
int k;
cin >> k;
for (int minute = 0; minute k; ++minute) {
int size = q.size();
for (int i = 0; i size; ++i) {
pairint,int > fr=q.front();//取到水管的位置坐标
int x = fr.first;
int y = fr.second;
q.pop();
for (int i2 = max(0,x - 1); i2 min(n,x + 1); i2++) { //四周灌溉
for (int j2 = max(0,y - 1); j2 min(m,y + 1); j2++) {
if (!irrigated[i2][j2]) {
irrigated[i2][j2] = true;
服务器托管网 q.push({i2, j2});
}
}
}
}
}
int count = 0;
for (int i = 0; i n; i++) {
for (int j = 0; j n; j++) {
if (irrigated[i][j]) {
服务器托管网count++;
}
}
}
cout count;
return 0;
}
也可使用map
#include
#include
using namespace std;
const int N = 200;
int main()
{
int n, m, t;
cin >> n >> m >> t;
bool irrigated[N][N];
mapint, int> q;
for (int i = 0; i t; i++) {
int r, c;
cin >> r >> c;
r--;
c--;
irrigated[r][c] = true;
q[r]=c;
}
int k;
cin >> k;
for (int minute = 0; minute k; ++minute) {
int size = q.size();
for (const auto& i:q) {
int x = i.first;
int y = i.second;
for (int i2 = max(0,x - 1); i2 min(n,x + 1); i2++) {
for (int j2 = max(0,y - 1); j2 min(m,y + 1); j2++) {
if (!irrigated[i2][j2]) {
irrigated[i2][j2] = true;
q[i2]=j2;
}
}
}
}
}
int count = 0;
for (int i = 0; i n; i++) {
for (int j = 0; j n; j++) {
if (irrigated[i][j]) {
count++;
}
}
}
cout count;
return 0;
}
[例3] 回文日期
评测系统
#include
#include
using namespace std;
bool huiwen(string s) {
if (s[0] == s[7] && s[1] == s[6] && s[2] == s[5] && s[3] == s[4]) {
return true;
}
else
return false;
}
bool er(string s) {
if (s[0] == s[2] && s[2] == s[5] && s[5] == s[7] && s[1] == s[3] && s[3] == s[4] && s[4] == s[6]) {
return true;
}
else
return false;
}
int pdday(int year, int month) {
if (month == 1 || month == 3 || month == 5 || month == 7 || month == 8 || month == 10 || month == 12)
return 31;
else if (month == 2) {
if (year % 4 == 0 && year % 100 != 0 || year % 400 == 0) {
return 29;
}
else
return 28;
}
else
return 30;
}
string zifuchuan(int year, int month, int day) {
string s=to_string(year);
if (month / 10 == 0) {
s += '0' + to_string(month);
}
else {
s += to_string(month);
}
if (day / 10 == 0) {
s += '0' + to_string(day);
}
else {
s += to_string(day);
}
return s;
}
int main()
{
int day, month, year;
int chushi;
int day1,day2;
cin >> chushi;
day1 = chushi % 10;
chushi = chushi / 10;
day2= chushi % 10;
chushi = chushi / 10;
day=day2 * 10 + day1;
day1 = chushi % 10;
chushi = chushi / 10;
day2 = chushi % 10;
chushi = chushi / 10;
month = day2 * 10 + day1;
year = chushi;
bool flag = 0;
bool pdhuiwen = 0;
bool pder = 0;
int i, j, k;
for (i = year;; i++) {
for (j = 1; j 12; j++) {
if (flag == 0) {
j = month;
}
for (k = 1; k pdday(year,month); k++) {
if (flag == 0) {
k = day+1;
flag = 1;
}
string s=zifuchuan(i, j, k);
if (pdhuiwen!=1&&huiwen(s)) {
cout s endl;
pdhuiwen = 1;
}
if (pder!=1&&er(s)) {
cout s;
pder = 1;
}
if (pdhuiwen == 1 && pder == 1) {
return 0;
}
}
}
}
}
[例4] 小蓝和小桥的挑战
评测系统
#include
#include
#include
using namespace std;
int main()
{
int n;
cin >> n;
while (n--) {
int m;
cin >> m;
vectorint> a(m);
for(int i=0;im;i++){
cin >> a[i];
}
int cnt = count(a.begin(), a.end(), 0); // #include
int sum = 0;
for (const auto& x : a) {
sum += x;
}
if (sum == 0) {
cout cnt + 1;
}
else
cout cntendl;
}
}
[例5] DNA序列修正
评测系统
#include
using namespace std;
bool pipei(char a, char b) {
return (a == 'A' && b == 'T') || (a == 'T' && b == 'A') || (a == 'C' && b == 'G') || (a == 'G' && b == 'C');
}
int main()
{
int n;
cin >> n;
string s1, s2;
cin >> s1 >> s2;
int cnt = 0;//计数
for (int i = 0; i n; ++i) {
if (!pipei(s1[i], s2[i])) {
cnt++;
for (int j = i + 1; j n; j++) {
if (!pipei(s1[j], s2[j])
&& pipei(s1[i], s2[j])
&& pipei(s1[j], s2[i])
) {
swap(s2[i], s2[j]);
break;
}
}
}
}
cout cnt;
}
[例6] 无尽的石头
评测系统
#include
using namespace std;
int ssum(int x) {
int sum = 0;
while (x) {
sum += x % 10;
x /= 10;
}
return sum;
}
int main()
{
int t;
cin >> t;
while (t--) {
int x = 0;
cin >> x;
int step = 0;
bool end = 0;
for (int i = 1; i x; i += ssum(i)) {
if (i == x) {
cout step endl;
end = 1;
}
step++;
}
if (end == 0) {
cout -1 endl;
}
}
}
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