代码源每日一题Div2
一个小整数
思路:
开始思路是找规律,for遍历每个数,然后插空什么的,结果问题就是有很多情况考虑不到.最后换了dfs解法,
首先判断数量最多的数的数量和其他数数量总和的关系
1.如果数量最多的数的数量大于其他数数量总和+1,那么一个隔一个插空都插不了,空不够,直接判失败
2.如果数量最多的数的数量等于其他数数量总和+1,这个时候就只能把最大的这个数推进用于排序的数组
3.如果数量最多的数的数量小于其他数数量总和+1,这个时候就是选合适的最小的数的时候了
最后把用于排序的数组遍历输出一下,结束
当然中间要处理一些特殊情况,比如如果只有一个零,那就在开始直接判断然后输出,不用进入dfs了
#define _CRT_SECURE_NO_WARNINGS
#include
int shuliang[10];
using namespace std;
int dfs(vector& a) {
int qiyv = 0, maxx = 0, maxshu = -1;
for (int e = 0; e < 10; e++) {
if (shuliang[e] > maxx) {
maxx = shuliang[e];
maxshu = e;
}
qiyv += shuliang[e];
}
if (qiyv == 0) {
return 1;
}
qiyv -= maxx;
if (qiyv + 1 < maxx) {
return 0;
}
if (qiyv + 1 == maxx) {
if (maxshu == 0 && a.size() == 0) {
return 0;
}
else {
a.push_back(maxshu);
shuliang[maxshu]--;
}
}
else {
int k1 = 0;
int pan1 = 0;
if (a.size() == 0) {
k1 = 1;
pan1 = -1;
}
else {
k1 = 0;
pan1 = a.back();
}
for (int e = k1; e < 10; e++) {
if (shuliang[e] > 0 && e !=pan1) {
a.push_back(e);
shuliang[e]--;
break;
}
}
}
dfs(a);
}
int main() {
int x = 1;
for (int e = 0; e < 10; e++) {
scanf("%d", &shuliang[e]);
if (e != 0 && shuliang[e] != 0) {
x = 0;
}
}
if (x&&shuliang[0]==1) {
printf("%d", 0);
return 0;
}
vectora;
if (dfs(a)) {
for (auto i : a) {
cout << i;
}
}
else {
cout << -1;
}
} 特殊正方形
思路:
开始string一个字符串为n个+,然后找规律,每一层把除了最外层里面的+变成.,之后在把除了最外层.里面的.变成+,每变一次输出一次同时用stack接收,之后把stack中的数据再输出出来就是完整的正方形
#include
using namespace std;
stackh;
int main()
{
int a;
cin >> a;
string m;
int panduan = 0;
int jishu = 0;
for (int e = 0; e < a; e++) {
m += "+";
}
for (int ii = 0; ii < a / 2; ii++) {
if (panduan == 0) {
for (int e = jishu; e < (a - jishu); e++) {
m[e] = '+';
}
panduan = 1;
}
else {
for (int e = jishu; e < (a - jishu); e++) {
m[e] = '.';
}
panduan = 0;
}
h.push(m);
for (int e = 0; e < m.size(); e++) {
cout << m[e];
}
cout << endl;
jishu++;
}
if (a % 2 != 0) {
if (panduan == 0) {
for (int e = jishu; e < (a - jishu); e++) {
m[e] = '+';
}
panduan = 1;
}
else {
for (int e = jishu; e < (a - jishu); e++) {
m[e] = '.';
}
panduan = 0;
}
for (int e = 0; e < m.size(); e++) {
cout << m[e];
}
cout << endl;
}
while (!h.empty()) {
cout << h.top() << endl;
h.pop();
}
} 走楼梯
思路:
用动态分布解决,建立数组dp[i][j],j分为三个,计算上两节的数量,分为dp[i+1][0],和dp[i+2][1],dp[i+2][2],进行计算.
#include
#include
using namespace std;
int dp[55][3];
int x[2] = { 1,2 };
int main() {
int aa;
cin >> aa;
dp[0][0] = 1;
for (int e = 1; e <=aa; e++) {//最外层
dp[e][0] = dp[e][0]+dp[e - 1][0] + dp[e - 1][1] + dp[e - 1][2];
if (e - 2 >= 0) {
dp[e][1] = dp[e][1] + dp[e - 2][0];
dp[e][2] = dp[e][2] + dp[e - 2][1];
}
}
/*for (int e = 0; e <= aa; e++) {
for (int i = 0; i < 3; i++) {
cout << dp[e][i];
}
cout << endl;
}*/
cout << dp[aa][0] + dp[aa][1] + dp[aa][2];
} 走路
思路:
之前先用dfs写的,逻辑到没啥问题,但是超时,全部超时,麻了,改成动态规划,开一个dp[i][j]二维数组,i表示走的次数,j的大小跟m+1一样大,目的就是统计看哪个点被走了,思路就是遍历每一个j如果是1,就加上ai或bi,一步一步下去,最后输出dp[n],结束,后面优化了下空间复杂度.
dfs算法
#include
using namespace std;
int n, m;
int hhh[105][2];
int panduan[105][3] = {0};
vectormm;
void dfs(int x,int y) {
if (y == n+1) {
if (m>=x) {
mm[x] = 1;
//cout << "JJJ" << endl;
}
return;
}
if (panduan[y][0] == 0) {
panduan[y][0] = 1;
dfs(x + hhh[y][0], y+1);
//cout << "hhh";
}
if (panduan[y][1] == 0) {
panduan[y][1] = 1;
dfs(x + hhh[y][1], y+1);
}
panduan[y][1] = 0;
panduan[y][0] = 0;
//cout << "kkk";
return;
}
int main() {
cin >> n >> m;
for (int e = 0; e <= m; e++) {
mm.push_back(0);
}
for (int e = 1; e <= n; e++) {
cin >> hhh[e][0] >> hhh[e][1];
}
dfs(0,1);
for (auto i : mm) {
cout << i;
}
} 动态规划
#include
using namespace std;
int shujv[105][2];
int dp[105][100005];
int main() {
int n, m;
cin >> n >> m;
for (int e = 1; e <= n; e++) {
cin >> shujv[e][0] >> shujv[e][1];
}
dp[0][0] = 1;
for (int e = 1; e <= n; e++) {
for (int i = 0; i <= m; i++) {
if (dp[e-1][i] == 1) {
if (i + shujv[e][0] <= m) {
dp[e][i + shujv[e][0]] = 1;
}
if (i + shujv[e][1] <= m) {
dp[e][i + shujv[e][1]] = 1;
}
}
}
}
for (int e = 0; e <= m; e++) {
cout << dp[n][e];
}
} 动态规划优化空间复杂度
#include
using namespace std;
int shujv[105][2];
vectordp(100005,0);
vectordp2(100005, 0);
vectordp3(100005, 0);
int main() {
int n, m;
cin >> n >> m;
for (int e = 1; e <= n; e++) {
cin >> shujv[e][0] >> shujv[e][1];
}
dp[0] = 1;
for (int e = 1; e <= n; e++) {
for (int i = 0; i <= m; i++) {
if (dp[i] == 1) {
if (i + shujv[e][0] <= m) {
dp2[i + shujv[e][0]] = 1;
}
if (i + shujv[e][1] <= m) {
dp2[i + shujv[e][1]] = 1;
}
}
}
dp= dp2;
dp2 =dp3;
}
for (int e = 0; e <= m; e++) {
cout < 简单分数统计
思路:
用了下map,就过了,挺突然的,出问题的地方在map遍历的时候不是按照插入数据的顺序,不过开始记录一下数据顺序就好了
#include
using namespace std;
int a, b, c;
maphhh;
mapren;
vectorren2;
int main() {
cin >> a >> b >> c;
for (int e = 1; e <= a; e++) {
string zanshi;
cin >> zanshi;
ren.insert(pair(zanshi, 1));
ren2.push_back(zanshi);
}
for (int e = 1; e <= b; e++) {
string zanshi;
int zanshi2;
cin >> zanshi >> zanshi2;
hhh[zanshi] = zanshi2;
}
for (int e = 1; e <= c; e++) {
string zanshi;
string zanshi1;
string zanshi2;
cin >> zanshi >> zanshi1 >> zanshi2;
if (ren[zanshi]>0) {
if (zanshi2 == "AC") {
ren[zanshi] += hhh[zanshi1];
}
}
else {
ren[zanshi] = -1;
}
}
for (int e = 0; e < ren2.size(); e++) {
map::iterator it;
for (it = ren.begin(); it != ren.end(); it++) {
if (it->first == ren2[e]) {
cout << it->first << " " << it->second - 1 << endl;
break;
}
}
}
} alice的德州扑克
思路:
单纯if,else嵌套了一下,判别点数相同就for遍历对比
#include
using namespace std;
maphhh;
int dianshu[6];
int main() {
int a[6], b[6];
cin >> a[1] >> a[2] >> a[3] >> a[4] >> a[5] >> b[1] >> b[2] >> b[3] >> b[4] >> b[5];
int panduansige = 0;
for (int e = 1; e <= 5; e++) {
int geshu = 0;
for (int i = 1; i <= 5; i++) {
if (a[e] == a[i]) {
geshu++;
}
}
dianshu[e] = geshu;
}
for (int e = 1; e <= 5; e++) {
if (dianshu[e] >= 4) {
panduansige = 1;
}
}
int sange = 0;
int liangge = 0;
for (int e = 1; e <= 5; e++) {
if (dianshu[e] == 3) {
sange++;
}
if (dianshu[e] == 2) {
liangge++;
}
}
/*for (int e = 1; e <= 5; e++) {
cout << b[e] << " ";
}*/
if (b[1]==b[2]&&b[2] == b[3]&&b[3] == b[4]&&b[4] == b[5]) {
//cout << "HH<" << endl;
if (a[5] == 14 && a[4] == 13 && a[3] == 12 && a[2] == 11 && a[1] == 10) {
cout << "ROYAL FLUSH";
}
else {
if (a[5] == a[4] + 1 && a[4] == a[3] + 1 && a[3] == a[2] + 1 && a[2] == a[1] + 1) {
cout << "STRAIGHT FLUSH";
}
else {
if (panduansige== 1) {
cout << "FOUR OF A KIND";
}
else {
if (sange == 3 && liangge == 2) {
cout << "FULL HOUSE";
}
else {
cout << "FLUSH";
}
}
}
}
}
else {
if (panduansige == 1) {
cout << "FOUR OF A KIND";
}
else {
if (sange == 3 && liangge == 2) {
cout << "FULL HOUSE";
}
else {
if (a[5] == a[4] + 1 && a[4] == a[3] + 1 && a[3] == a[2] + 1 && a[2] == a[1] + 1) {
cout << "STRAIGHT";
}
else {
cout << "FOLD";
}
}
}
}
} 订单编号
思路:
开始不知道这个题考查什么,直接列了map,查询,结果123过了,456超时;
最后把点搜索换成区间搜索,还是超时没过,迷惑,最后发现是cin和cout的问题,换成scanf和printf就好了
map超时代码:
#include
using namespace std;
mappanduan;
int main() {
int a;
cin >> a;
for (int e = 0; e < a; e++) {
int x;
cin >> x;
if (panduan[x] == 0) {
panduan[x] = 1;
cout << x<<" ";
}
else {
for (int e = x + 1;; e++) {
if (panduan[e] == 0) {
panduan[e] = 1;
cout << e << " ";
break;
}
}
}
}
} 区间搜索用scanf 和printf代码:
#include
using namespace std;
set>hhh;
inline void insert(int x, int y) {
if (x < y) {
return;
}
hhh.insert(make_pair(x, y));
}
int main() {
int a;
scanf("%d", &a);
insert(2e9, 1);
for (int e = 0; e < a; e++) {
int x;
scanf("%d", &x);
auto its = hhh.lower_bound(make_pair(x, 0));
if (its->second <= x) {
printf("%d ", x);
insert(its->first, x + 1);
insert(x-1,its->second);
hhh.erase(its);
}
else {
printf("%d ", its->second);
insert(its->first, its->second + 1);
hhh.erase(its);
}
}
} 任务分配
思路:
用动态规划,这个和之前有个叫采药的题比较像,思路就是遍历时间数组,每次到一个点先看是这个点的数大还是前一个点的数大,如果是前一个点数大那就把前一个点的数放到这个点,如果这个点和某个事件的开始时间相符合,那就判断这个事件是否值得执行.
#define _CRT_SECURE_NO_WARNINGS
#include
using namespace std;
int n;
struct s {
int st, ed,w;
};
vectora;
int main() {
int maxx = 0;
cin >> n;
for (int e = 0; e < n; e++) {
s x;
cin >> x.st >> x.ed >> x.w;
a.push_back(x);
if (x.ed > maxx) {
maxx = x.ed;
}
}
vectorpd(maxx+1);
for (int e = 1; e 路径计数
思路:
开始没多想,直接bfs,结果超时
#define _CRT_SECURE_NO_WARNINGS
#include
using namespace std;
int tu[105][105];
int panduan[105][105] = { 0 };
queue>a;
inline void push(int x, int y) {
a.push(make_pair(x, y));
}
int main() {
int x;
long long cishu=0;
cin >> x;
for (int e = 1; e <= x; e++) {
for (int i = 1; i <= x; i++) {
cin >> tu[e][i];
}
}
a.push(make_pair(1, 1));
panduan[1][1] = 1;
while (!a.empty()) {
int x1 = a.front().first;
int y1 = a.front().second;
a.pop();
if (x1 == x && y1 == x) {
cishu++;
cishu = cishu%1000000007;
continue;
}
if (x1+1<=x) {
if (tu[x1 + 1][y1] == 1) {
push(x1 + 1, y1);
if (x1 + 1 != x && y1 != x) {
panduan[x1 + 1][y1] = 1;
}
}
}
if (y1+1<=x) {
if (tu[x1][y1 + 1] == 1 ) {
push(x1, y1+1);
if (x1!= x && y1+1!= x) {
panduan[x1][y1 + 1] = 1;
}
}
}
}
cout << cishu;
}
换成动态规划
dp[i][j]=dp[i-1][j]+dp[i][j-1]过了
#define _CRT_SECURE_NO_WARNINGS
#include
using namespace std;
int tu[105][105];
long long panduan[105][105];
queue>a;
inline void push(int x, int y) {
a.push(make_pair(x, y));
}
int main() {
int x;
long long cishu = 0;
cin >> x;
for (int e = 1; e <= x; e++) {
for (int i = 1; i <= x; i++) {
cin >> tu[e][i];
}
}
panduan[1][1] = 1;
for (int e = 1; e <= x; e++) {
for (int i = 1; i <= x; i++) {
panduan[e][i] += panduan[e - 1][i] +panduan[e][i - 1];
panduan[e][i] = panduan[e][i] % 1000000007;
if (tu[e][i] == 0) {
panduan[e][i] = 0;
}
}
}
cout << panduan[x][x];
}
顺便温习了一下dfs,当然,也是超时
#define _CRT_SECURE_NO_WARNINGS
#include
using namespace std;
int tu[105][105];
int pan[105][105] = {0};
int x1;
long long jishu=0;
void dfs(int x, int y) {
if (x == x1 && y == x1) {
jishu++;
jishu = jishu % 1000000007;
return;
}
if (pan[x + 1][y] == 0&&tu[x+1][y]==1) {
pan[x + 1][y] = 1;
dfs(x + 1, y);
}
if (pan[x][y + 1] == 0 && tu[x][y+1] == 1) {
pan[x][y + 1] = 1;
dfs(x, y + 1);
}
pan[x][y + 1] = 0;
pan[x + 1][y] = 0;
return;
}
int main() {
long long cishu = 0;
cin >> x1;
for (int e = 1; e <= x1; e++) {
for (int i = 1; i <= x1; i++) {
cin >> tu[e][i];
}
}
dfs(1, 1);
cout << jishu;
}
最大和上升子序列
思路:
用dp从左到右扫一遍
#define _CRT_SECURE_NO_WARNINGS
#include
using namespace std;
int main() {
int n;
cin >> n;
int a[10005];
int f[10005];
for (int e = 1; e <= n; e++) {
cin >> a[e];
f[e] = a[e];
}
for (int e = 1; e <= n; e++) {
for (int i = 1; i < e; i++) {
if (a[e] > a[i]) {
f[e] = max(f[e], f[i] + a[e]);
}
}
}
int ans = 0;
for (int e = 1; e <= n; e++) {
ans = max(ans, f[e]);
}
cout << ans;
}