算法
十进制转化任意进制
#include <iostream>
#include<cstring>
#include <algorithm>//reverse()函数使用所需头文件
using namespace std;
string convertToBase8(int num) {
string ret;
bool isNegative = false;
if (num < 0)
{
num = num * (-1);
isNegative = true;
}
if (num == 0)
{
return "0";
}
while (num > 0)
{
ret.push_back(num % 8 + '0');
num = num / 8;
}
if (isNegative)
{
ret.push_back('-');
}
//使用了string类中的一些操作
reverse(ret.begin(), ret.end());//reverse()会将区间[beg,end)内的元素全部逆序
return ret;
}
int main() {
int n = 0 ;
cin >> n;
cout << convertToBase8(n);
}
全排列
1.问题解决输入遇到逗号停止
下面这个只能单个逗号才会停止
string a;
string list[10];
int c = 0;
while (cin>>a) {
if (a == ",") break;
else {
list[c] = a;
c++;
}
}
#include<iostream>
#include<cstring>
using namespace std;
inline void Swap(string& a, string& b) {
string temp = a;
a = b;
b = temp;
}
void Perm(string list[], int k, int m) {
if (k == m) {
for (int i = 0; i <= m; i++) cout << list[i];
cout << " ";
}
else {
for (int i = k; i <= m; i++) {
Swap(list[k], list[i]);
Perm(list, k + 1, m);
Swap(list[k], list[i]);
}
}
}
int main() {
char a;
string list[10];
int c = 0;
while (cin>>a) {
if (a == ',') break;
else {
list[c] = a;
c++;
}
}
Perm(list, 0, c);
}
//借助al的头文件解决字典顺序输入的全排列
#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;
void Perm(string list[], int m) {
do {
for (int i = 0; i <= m; i++) {
cout << list[i];
}
cout << " ";
} while (next_permutation(list, list + m));
}
int main() {
char a;
string list[10];
int c = 0;
while (cin >> a) {
if (a == ',') break;
else {
list[c] = a;
c++;
}
}
sort(list, list + c);
Perm(list, c );
}
第k小
#include<iostream>
using namespace std;
int Partition(int a[], int l, int r)
{
int i = l, j = r + 1;
int x = a[l];
while (true)
{
while (a[++i] < x && i <= r);
while (a[--j] > x);
if (i >= j)
break;
int m;
m = a[i];
a[i] = a[j];
a[j] = m;
}
a[l] = a[j];
a[j] = x;
return j;
}
int select(int a[], int left, int right, int k)
{
if (left == right)
return a[left];
int i = Partition(a, left, right);
int j = i - left + 1;
if (k <= j)
return select(a, left, i, k);
else
return select(a, i + 1, right, k - j);
}
int main()
{
int m, k;
cin >> m >> k;
int* p = new int[m];
for (int i = 0; i < m; i++)
cin >> p[i];
int h = select(p, 0, m - 1, k);
cout << h;
return 0;
}
//这个算法没有使用以五个为一组
快速幂及取余
int PowerMod(int a, int b, int c)
{
int ans = 1;
a = a % c;
while (b > 0）
{
if (b%2 == 1)
ans = (ans * a) % c;
b = b / 2;
a = (a * a) % c;
}
return ans
}//快速幂取余，a的b次方对c取余
#include<iostream>
#include<algorithm>
using namespace std;
int qpow(int a, int n)
{
if (n == 0)
return 1;
else if (n % 2 == 1)
return qpow(a, n - 1) * a;
else
{
int temp = qpow(a, n / 2);
return temp * temp;
}
}
int zz(int aa) {
int num = 0;
for (int i = 0; i <= aa; i++) {
num += qpow(i, i);
}
return num % 100000007 ;
}
int main() {
int x,y,z;
cin >> x >> y >> z;
cout << zz(x) << endl;
cout << zz(y) << endl;
cout << zz(z) << endl;
}
#include<iostream>
using namespace std;
const long long N = 100000007;
#define lx long long
lx m;
lx hs(lx n,lx m,lx k)
{
lx ans=1%k;
while(m)
{
if (m&1)
ans=ans*n%k;
n=n*n%k;
m>>=1;
}
return (ans%k);
}
int main(){
int m;
while(cin>>m){
lx res=1;
for(int i=1;i<=m;i++){
lx t = hs(i,i,N)%N;
res=(res+t)%N;
}
cout<<res%N<<endl;
}
}
Joseph问题
#include<iostream>
using namespace std;
int vis[16] = { 0 };
int fond(int k)
{
if (vis[k]) //如果k是曾经出现过的，已经计算了答案的，就直接输出；
return vis[k];
else
{
for (int i = k + 1;; i++) //从k+1开始找
{
int sum = k * 2, flag = 0; //sum为所剩的人数，flag来标记一个数是否为后k个数，如果flag==1，这个i不用继续找下去了；
for (int j = i; flag == 0; j += i - 1)
{
if (j > sum)
j = j % sum ? j % sum : sum;
if (j <= k)
break;
else
sum--;
if (sum == k)
flag = 1;
}
if (flag)
{
vis[k] = i;
return vis[k];
}
}
}
}
int main()
{
int n;
while (cin >> n && n)
cout << fond(n) << endl;
return 0;
}
火柴
#include<iostream>
#include<stdio.h>
#include<algorithm>
using namespace std;
int main() {
int n=1, val=0;
while (n) {
cin >> n;
int a = 0;
for (int i = 0; i < n; i++) {
cin >> val;
a ^= val;
}
if (a != 0) {
cout << "Yes" << endl;
}
else {
cout << "No" << endl;
}
b--;
}
}
Factstone Benchmark
//位数的意思是：比如4位可以表示2^4=16个0到15 16个整数
#include <iostream>
#include <cmath>
using namespace std;
int main()
{
int y; //y表示年份
int w; //w是指数，2^w表示位数（字长），即len
int n; //n为FactStone评级
int len; //len是字长（位数），
//如w=4，即len=2^4=16，位数为16位，可以表示2^16个整数（0~2^16-1)
while (1) //不断执行循环，跳出条件为输入0，下面写出
{
cin >> y;
if (y == 0) //程序结束执行的条件
{
break;
}
int w = (y / 10) - 194; //w分别取2，3，4，5……
len = pow(2, w); //求出位数（字长）len
n = 1;
double result = 0;
while (result < len*log(2)) //这是纸上推出来的
{
result = result + log(n); //乘法转换为了加法
n++;
}
cout << n - 2 << endl; //关于这里为什么是n-2，请看下面解释
}
return 0;
}
内部收益率
#include<iostream>
#include<cmath>
#include <stdio.h>
using namespace std;
double polyval(int*x,int y, double z)
{
double val=0.0;
for (int i = 0; i <= y; i++) {
val += x[i] / pow(z+1,i);
}
return val;
}
double ddd(double &c) {
int T;
cin >> T;
int CF[30] = { 0 };
if (T > 0) {
for (int i = 0; i <= T; i++) {
cin >> CF[i];
}
double eps = 0.01;
double x2 = 1, x1 = -1;
double x = (x2 - x1) / 2;
while ((x2 - x1) >= eps)
{
if (polyval(CF, T, x1) * polyval(CF, T, x) > 0)
{
x1 = x;
x = (x1 + x2) / 2;
}
else if (polyval(CF, T, x1) * polyval(CF, T, x) == 0)
{
break;
}
else
{
x2 = x;
x = (x1 + x2) / 2;
}
}
c = x;
return c;
}
else {
c = -2;
return c;
}
}//多项式利用二分法进行求解EDGE
int main() {
double c1 = 0.0, c2 = 0.0,c3 = 0.0;
ddd(c1);
ddd(c2);
ddd(c3);
if (c1 != -2) {
printf("%.2lf", c1);
cout << endl;
}
if (c2 != -2) {
printf("%.2lf", c2);
}
if (c3 != -2) {
printf("%.2lf", c3);
}
return 0;
}
跳台阶
#include<iostream>
using namespace std;
int jump(int n) {
if (n == 1 || n == 0) {
return 1;
}
else {
int c = (jump(n - 2) + jump(n - 1)) % 1000000007;
return c;
}
}
int main() {
int n;
while (cin >> n) {
cout << jump(n) << endl;
}
return 0;
}//递归解法，时间会超限
最长递减子序列
#include<bits/stdc++.h>
using namespace std;
int main()
{
int a[1000],b[1000],c[1000]={0}, res = 1;
int n;
cin >> n;
for (int i = 1; i <= n; i++)
cin >> a[i];
for (int i = 1; i <= n; i++) {
for (int j = 1; j < i; j++) {
if (a[i] < a[j] && b[i] < (b[j] + 1)) {
b[i] = b[j] + 1;
c[i] = j;//核心代码
}
}
if (b[i] > b[res]) {
res = i;
}
}
stack<int> Q;
for (int i = res; i != 0; i = c[i])
Q.push(a[i]);
cout << Q.top();
Q.pop();
while (!Q.empty())
{
cout << " " << Q.top();
Q.pop();
}
return 0;
}
#include<iostream>
#include<algorithm>
using namespace std;
int dp[502][502], map[502][502];
int x[4] = { 0,0,-1,1 }, y[4] = { -1,1,0,0 };//移动方向
int r, c;//r行c列的二维矩阵
int search(int i, int j) {
if (dp[i][j] > 1) {
return dp[i][j];
}
for (int k = 0; k < 4; k++) {
if (i + x[k] > 0 && i + x[k] <= r && j + y[k] > 0 && j + y[k] <= c && map[i][j] > map[i + x[k]][j + y[k]])
dp[i][j] = max(dp[i][j], search(i + x[k], j + y[k]) + 1);
}
return dp[i][j];
}
int main() {
cin >> r >> c;
for (int i = 1; i <= r; i++) {
for (int j = 1; j <= c; j++) {
dp[i][j] = 1;
cin >> map[i][j];
}
}
int cc = 0;
for (int i = 1; i <= r; i++) {
for (int j = 1; j <= c; j++) {
dp[i][j] = search(i, j);
cc = max(cc, dp[i][j]);
}
}
cout << cc << endl;
return 0;
}//好理解一点动态规划
矩阵滑雪场
#include<bits/stdc++.h>
using namespace std;
int n, m, a[1001][1001], ans, fx[4][2] = { 0,1,1,0,-1,0,0,-1 }, step[1001][1001];
int dfs(int x, int y) {
int mmax = 0;
if (step[x][y] != 0)return step[x][y];
step[x][y] = 1;//最小的点为1
for (int i = 0; i < 4; i++) {
int xx = x + fx[i][0];
int yy = y + fx[i][1];
if (xx >= 1 && xx <= n && yy >= 1 && yy <= m && a[xx][yy] < a[x][y])step[x][y] = max(step[x][y], dfs(xx, yy) + 1);
}
return step[x][y];
}
int main() {
cin >> n >> m;
for (int i = 1; i <= n; i++)
for (int j = 1; j <= m; j++)cin >> a[i][j], step[i][j] = 0;
for (int i = 1; i <= n; i++)
for (int j = 1; j <= m; j++) {
ans = max(ans, dfs(i, j));
}
cout << ans;
return 0;
}
最长子序列
#include<iostream>
using namespace std;
int main() {
int n;
cin >> n;
int a[1000], b = 0;
for (int i = 0; i < n; i++) {
cin >> a[i];
}
int sum = 0;
for (int i = 0; i < n; i++) {
if (b > 0) {
b += a[i];
}//b>0证明前面子序列是正整数并且予以sum保存
else {
b = a[i];
}
if (b > sum) {
sum = b;
}
}
cout << sum;
}//仅要求返回子序列的最大和
元素整除问题
#include<iostream>
using namespace std;
int main() {
int a[20] = { 0 };
for (int i = 0; i < 20; i++) {
cin >> a[i];
}
int b[20] = { 0 };
for (int i = 0; i < 20; i++) {
int c = a[i];
for (int j = i + 1; j < 20; j++) {
if (a[j] % c == 0) {
b[j] = 1;
}
}
}
for (int j = 0; j < 20; j++) {
if (b[j] == 1) {
cout << a[j] << endl;
}
}
}#include<iostream>
using namespace std;
int main() {
int a[20] = { 0 };
for (int i = 0; i < 20; i++) {
cin >> a[i];
}
int b[20] = { 0 };
for (int i = 0; i < 20; i++) {
int c = a[i];
for (int j = i + 1; j < 20; j++) {
if (a[j] % c == 0) {
b[j] = 1;
}
}
}
for (int j = 0; j < 20; j++) {
if (b[j] == 1) {
cout << a[j] << endl;
}
}
}//遍历数组将能被整除的数用另一数组作为标记，最终输出只需输出被标记位置元素
渊子赛马
#include<iostream>
#include<algorithm>
using namespace std;
int main() {
int n;
int a[1005], b[1005];
while (cin >> n && n) {
for (int i = 0; i < n; i++) cin >> a[i];
for (int i = 0; i < n; i++) cin >> b[i];
sort(a, a + n);
sort(b, b + n);
int count = 0;
int A_min = 0, A_max = n - 1;
int B_min = 0, B_max = n - 1;
while (A_min <= A_max) {
if (a[A_max] > b[B_max]) {
count++;
A_max--;
B_max--;
}
else if (a[A_min] > b[B_min]) {
count++;
A_min++;
B_min++;
}
else {
count--;
A_min++;
B_max--;
}
}
if (count > n / 2) cout << "YES" ;
else cout << "NO" ;
}
return 0;
}//这个没过
#include <iostream>
#include <string.h>
#include <cstdio>
using namespace std;
int a[1005], b[1005];
void paixu(int* c, int n)
{
int t, i = 0, j = 0;
for (i = 0; i < n - 1; i++)
{
for (j = 0; j < n - i - 1; j++)
{
if (c[j] > c[j + 1])
{
t = c[j];
c[j] = c[j + 1];
c[j + 1] = t;
}
}
}
}
int main()
{
int n, i, j, win, lose;
while (1)
{
cin >> n;
if (n == 0)
{
break;
}
for (i = 0; i < n; i++)
{
cin >> a[i];
}
for (i = 0; i < n; i++)
{
cin >> b[i];
}
paixu(a, n);
paixu(b, n);
for (i = 0, j = 0, win = 0, lose = 0; i < n; i++)
{
if (a[i] > b[j])
{
j++;
win++;
}
else
{
lose++;
}
if (win > n / 2 || lose > n / 2)
break;
}
if (win > n / 2)
{
cout << "YES" << endl;
}
else
{
cout << "NO" << endl;
}
}
return 0;
}
货币背包问题（这个有格式化输出）
#include<bits/stdc++.h>
using namespace std;
long long int dp[40005];
int coin[] = { 5,10,20,50,100,200,500,1000,2000,5000,10000 };
int main(){
dp[0] = 1;
for (int i = 0; i <= 10; i++)
for (int j = coin[i]; j <= 30000; j++)
dp[j] += dp[j - coin[i]];
double n;
while (cin >> n && n) {
int m = (int)(n * 100 + 0.0001);
printf("%6.2f%17lld\n", n, dp[m]);
}
return 0;
}
沙子质量
#include<iostream>
#include<algorithm>
using namespace std;
const int N = 1000;
int x[N + 1], dp[N + 1][N + 1], s[N + 1];
int n,sum;
int main() {
while (cin >> n) {
for (int i = 1; i <= n; i++) {
cin >> x[i];
s[i] = s[i - 1] + x[i];
}
for (int i = 1; i <= n; i++) {
dp[i][i] = x[i];
}
for (int r = 2; r <= n; r++) {
for (int i = 1; i <= n - r + 1; i++) {
int j = i + r - 1;
dp[i][j] = dp[i + 1][j]+dp[i][j-1];
for (int k = i + 1; k < j; k++) {
int t = min(dp[i][j],dp[i][k] + dp[k + 1][j]+s[j]-s[i-1]);
if (t < dp[i][j]) {
dp[i][j] = t;
}
}
}
}
cout << dp[1][n] << endl;
}
return 0;
}//自己代码，没过待查证
e//沙子的质量(跟矩阵的连乘问题类似，都是自底向上）
#include<iostream>
#include<cstring>
using namespace std;
int f[310][310];
int a[310];
int main(){
memset(f,0x3f,sizeof f);//将矩阵初始化，各元素无穷大
int n;
cin>>n;
for (int i = 0; i < n; ++i){
cin>>a[i];
}
for (int i = 0; i < n; i++)
f[i][i] = 0;
for (int i = 1; i < n; ++i)
a[i] = a[i] + a[i - 1];
for (int k = 2; k <= n; ++k){
for (int i = 0; i + k - 1 < n; ++i){
int j = i + k - 1;
for (int z = i; z < j; ++z){
f[i][j] = min(f[i][j],(f[i][z] + f[z+1][j] + a[j] - a[i - 1]));
}
}
}
cout<<f[0][n - 1]<<endl;
return 0;
}对比吧
最长公共子序列
# include <iostream>
# include <cstring>
using namespace std;
int f[2010][2010];
char a[2010],b[2010];
int max(int a,int b)
{
if (a>=b) return a;
return b;
}
int main()
{
int i,j,len1,len2;
cin>>a>>b;
len1=strlen(a);
len2=strlen(b);
for(i=1;i<=len1;i++)
{
for(j=1;j<=len2;j++)
{
if (a[i-1]==b[j-1]) f[i][j]=f[i-1][j-1]+1;
else f[i][j]=max(f[i-1][j],f[i][j-1]);
}
}
cout<<f[len1][len2];
return 0;
}
三角形权
#include<bits/stdc++.h>
using namespace std;
int main() {
int n;
cin >> n;
int a[26][26],dp[26][26]={0}//数组最好还是初始化，尤其大的时候;
for (int i = 0; i < n; i++) {//C中有memset库函数就是用来初始化
for (int j = 0; j <= i; j++) {
cin >> a[i][j];
}
}
for (int i = n - 1; i >= 0; i--)
for (int j = 0; j <= i; j++)
dp[i][j] = max(dp[i + 1][j], dp[i + 1][j + 1]) + a[i][j];
cout<<dp[0][0];
return 0;
}
跳跃游戏2
#include<bits/stdc++.h>
using namespace std;
int main() {
int n;
cin >> n;
int a[101];
for (int i = 0; i < n; i++) {
cin >> a[i];
}
int i = 0;
int count = 0;
if (n == 1)
{
cout << 0 << endl;
return 0;
}//只有一个直接return 0
while (true)
{
if (i + a[i] >= n - 1)
{
count++;
break;
}//终止条件
int maxNum = -1;
int maxIndex;
for (int j = i + 1; j <= i + a[i]; j++)
if (a[j] >= maxNum)
{
maxNum = a[j];
maxIndex = j;
}
i = maxIndex; // 跳跃到最合适的位置
count++;
}
cout << count << endl;
return 0;
}//无法AC通过，贪心算法，存在问题：无法回跳完的情况
homework
#include<bits/stdc++.h>
using namespace std;
struct qwe {
double t;
double v;
}a[21];
bool cmp(qwe x, qwe y)
{
return x.v / x.t > y.v / y.t;
}
int main() {
qwe a[21];
int M, N;
while (cin >> M >> N && M && N) {
for (int i = 0; i < M; i++) {
cin >> a[i].t >> a[i].v;
}
sort(a, a + M, cmp);//按照价值/时间的值进行排序
double sum = 0;
int i;
for (i = 0; i < M; i++)
{
if (a[i].t <= N) // 如果时间够的话
{
N -= a[i].t;
sum += a[i].v;
}
else
break;
}
if (i < M) // 如果能够完成所有题, 就不用进入这个循环了
sum += 1.0 * a[i].v / a[i].t * N; // 剩余的时间能完成多少价值.
printf("%.2lf\n", sum);
}
}//一个背包问题
法师康的工人 （贪心）
#include<iostream>
#include<algorithm>
using namespace std;
struct Worktime {
int start;
int end;
};
bool cmp(Worktime a,Worktime b)
{
if (a.start != b.start)
return a.start < b.start;
return a.end < b.end;
}//排序规定
int main()
{
int N;
cin >> N;
Worktime* p = new Worktime[N];
for (int i = 0; i < N; i++)
{
int x;
cin >> x;
p[i].start = x;
int y;
cin >> y;
p[i].end = y;
}
sort(p, p + N, cmp);
int max1 = p[0].end - p[0].start;
int max2 = 0;
int start = p[0].start;
int end = p[0].end;
for (int i = 0; i < N - 1; i += 1)
{
if (end >= p[i + 1].start)
{
end = max( p[i + 1].end , end);
max1 = max( end - start , max1 );
}
else
{
start = p[i + 1].start;
max2 = max( p[i + 1].start - end , max2);
end = p[i + 1].end;
}
}
cout << max1 << ' ' << max2 << endl;
}
三值问题
#include<iostream>
using namespace std;
int a[1001], cnt[4];
int main()
{
int n, i;
cin>>n;
for (i = 1; i <= n; i++)
{
cin>>a[i];
cnt[a[i]]++;
}
int sum1 = 0, sum2 = 0, sum3 = 0;
for (i = 1; i <= cnt[1]; i++)
{
if (a[i] != 1)
sum1++;
}
for (i = cnt[1] + 1; i <= cnt[1] + cnt[2]; i++)
{
if (a[i] == 3)
sum2++;
}
for (i = cnt[1] + cnt[2] + 1; i <= n; i++)
{
if (a[i] == 2)
sum3++;
}
cout<<sum1 + (sum2 > sum3 ? sum2 : sum3);
return 0;
}
思路：首先分析，无论交换前的数据如何，交换后一定可以分为三部分，第一部分是数据“1”，第二部分是数据“2”，第三部分是数据“3”，由于“1”永远在最前面，所以我门可以先固定第一部分，然后讨论第二、三部分
（1）首先找出数据中 “1”的个数sum1，即完成第一部分排序需要的步数
（2）然后要做的就只有 将第二部分中的“3”交换到第三部分、或者将第三部分的“2”交换到第二部分，分别记为sum2和sum3
（3）取sum1+max(sum2,sum3),至于为什么要取sum2和sum3之间的最大值，是因为如果取最小值，那么必定会有“2”遗留在第三部分或者“3”遗留在第二部分，取决于两者到底谁小
区间包含问题
#include <iostream>
#include <algorithm>
using namespace std;
int m, n;
struct node {
int s, e;
}a[100010], b[100010];
int cmp(node a, node b) {
if (a.e == b.e)
return a.s < b.s;
return a.e < b.e;
}
int fun1(node s) {
int num1 = 0, num2 = 0;
int left = 1;
int right = m;
int mid;
while (left < right) {
mid = (left + right) / 2;
if (a[mid].e == s.e)
return mid;
else if (a[mid].e > s.e)
right = mid - 1;
else
left = mid + 1;
}
return mid;
}//快排方法
int main() {
while (cin >> m >> n)
{
for (int i = 1; i <= m; i++)
cin >> a[i].s >> a[i].e;
sort(a + 1, a + 1 + m, cmp);
for (int j = 1; j <= n; j++)
{
int num = 0;
cin >> b[j].s >> b[j].e;
int t = fun1(b[j]);
for (int i = 1; i <= t + 1; i++)
if (a[i].s >= b[j].s && a[i].e <= b[j].e)
{
num++;
b[j].s = a[i].e;
}
cout << num << endl;
}
}
return 0;
}
The Hardest Problem Ever
#include<bits/stdc++.h>
using namespace std;
int main()
{
int i, n;
char a[205], str[205];
char s[27] = { 'V','W','X','Y','Z','A','B','C','D','E','F','G','H','I','J','K','L','M','N','O','P','Q','R','S','T','U' };//字符串空间最好开大点
while (cin.getline(a, 205))//多组输入
{
if (strcmp(a, "ENDOFINPUT") == 0)//如果输入的字符串a，与ENDOFINPUT相同，那么就跳出循环
{
break;
}
if (strcmp(a, "START") == 0)//如果输入的字符串a，与START相同，那么就执行以下程序
{
cin.getline(str, 205);
n = strlen(str);
for (i = 0; i < n; i++)
{
if (str[i] >= 'A' && str[i] <= 'Z')
{
str[i] = s[str[i] - 'A'];//减去A对应的ASCII码就是相应的明文
}
}
cout<<str;
}
}
return 0;
}
8皇后问题
#include<bits/stdc++.h>
using namespace std;
const int n = 8;
int tmp, ans;
int x[10];
int chess[10][10];
bool notDanger(int r, int c) {
for (int i = 0; i < r; i++) {
if (x[i] == c) return false;
if (abs(r - i) == abs(c - x[i])) return false;
}
return true;
}
void queen(int k) {//放入第k行的皇后
if (k == n) {
ans = max(ans, tmp);
return; }//所有行都尝试完毕，即求出了相应一组解
for (int i = 0; i < n; i++) {
if (notDanger(k, i)) {//没有冲突，尝试下一行
x[k] = i;
tmp += chess[k][i];
queen(k + 1);
x[k] = -1;//注意下层递归结束后及时更新相应变量值
tmp -= chess[k][i];
}
}
}
int main() {
int k;
cin>>k;
while (k--) {
tmp = ans = 0;
x[10] = { 0 };
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++)
cin>>chess[i][j];
}
queen(0);
cout << ans;
}
return 0;
}
无脑博士的试管们
#include<bits/stdc++.h>
using namespace std;
int a, c, b, A, B, C;//a,b,c代表此时A,B,C容器有多少水
int vis[25][25];//用VIS数组来表示每次变换时的A,B杯的状态
void dfs(int a, int b, int c) {
vis[a][b] = 1;//状态A杯为a,B杯为b
if (a < A) //当A容器没有装满时的情况
{
if (c >= A - a && vis[A][b] == 0)dfs(A, b, c - A + a);//从c倒入a中，C容器里的大于A剩余空间，可装满
if (c < A - a && vis[a + c][b] == 0)dfs(a + c, b, 0); //c倒入a中，C容器里的液体小于A的剩余空间，以下也是
if (b >= A - a && vis[A][b - A + a] == 0)dfs(A, b - A + a, c);
if (b < A - a && vis[a + b][0] == 0)dfs(a + b, 0, c);
}
if (b < B) {
if (c >= B - b && vis[a][B] == 0)dfs(a, B, c - B + b);
if (c < B - b && vis[a][b + c] == 0)dfs(a, b + c, 0);//从c倒入b
if (a >= B - b && vis[a - B + b][B] == 0)dfs(a - B + b, B, c);//从a倒入b
if (a < B - b && vis[0][a + b] == 0)dfs(0, a + b, c);
}
if (c < C) {
if (a >= C - c && vis[a - C + c][b] == 0)dfs(a - C + c, b, C);
if (a < C - c && vis[0][b] == 0)dfs(0, b, a + c);;//从a倒入c
if (b >= C - c && vis[a][b - C + c] == 0)dfs(a, b - C + c, C);
if (b < C - c && vis[a][0] == 0)dfs(a, 0, b + c);//从b倒入c
}
}
int main() {
cin >> A >> B >> C;
vis[25][25] = { 0 };
dfs(0, 0, C);
int flag = 1;
for (int i = B; i >= 0; i--) {
if (vis[0][i]) {
if (flag) flag = 0;//这个flag的存在可以去出最后多出来的空格
else cout << " ";
cout << C - i;
}
}
return 0;
}
图的m着色问题
#include<iostream>
using namespace std;
int n, k, m, sum = 0;
int color[20000];
int connect[2000][2000];
void dfs(int d)
{
if (d == n + 1)//所有顶点全部着色
{
sum++;//着色方案
return;
}
for (int i = 1; i <= m; i++)//循环颜色
{
bool flag = 1;
for (int j = 1; j <= n; j++)
{
if (connect[d][j] && color[j] == i)
{
flag = 0;
break;
}
}
if (flag)
{
color[d] = i;//尝试分配颜色；
dfs(d + 1);
color[d] = 0;//不行则回退；
}
}
}
int main()
{
cin >> n >> k >> m;
for (int i = 0; i < k; i++)
{
int tmp1, tmp2;
cin >> tmp1 >> tmp2;
connect[tmp1][tmp2] = 1;
connect[tmp2][tmp1] = 1;
}
dfs(1);
cout << sum;
return 0;
}
组合运算式
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
char s[20];
int n;
void DFS(int pos, int sum, int last)///三个参数：当前位置，不含当前位置的和，当前位置上的数
{
int i;
if (pos == n)
{
if (sum + last == 0)
{
cout<<s;
cout << endl;
}
return;
}
s[pos * 2 - 1] = ' '; //搜索‘ ’号
if (last > 0)
{
DFS(pos + 1, sum, last * 10 + pos + 1);
}
else
{
DFS(pos + 1, sum, last * 10 - pos - 1);
}
s[pos * 2 - 1] = '+'; //搜索‘+’号
DFS(pos + 1, sum + last, pos + 1);
s[pos * 2 - 1] = '-'; //搜索‘-’号
DFS(pos + 1, sum + last, -(pos + 1));
}
int main()
{
int i;
cin>>n;
for (i = 0; i < n; i++)
{
s[i * 2] = i + '1';
}
DFS(1, 0, 1);
return 0;
}
凯撒加密算法
#include<iostream>
#include<cstring>
using namespace std;
int main() {
int T,s;
cin >> T;
while (T--) {
string p;
cin >> p;
cin >> s;
s %= 26;
for (int i = 0; i < p.length(); i++) {
if (p[i] >= 'A' && p[i] <= 'Z')
p[i] = 'A' + (p[i] - 'A' - s + 26) % 26;
if (p[i] >= 'a' && p[i] <= 'z')
p[i] = 'a' + (p[i] - 'a' - s + 26) % 26;
}
cout << p << endl;
}
}
简单的密码
#include<bits/stdc++.h>
using namespace std;
int n, f[35] = { 0,0,0,1 };
int main()
{
for (int i = 4; i <= 30; i++)
f[i] = 2 * f[i - 1] + (1 << (i - 4)) - f[i - 4];
while (cin >> n)
cout << f[n] << endl;
return 0;
}
数据加密
#include<bits/stdc++.h>
using namespace std;
int n, tip, top;
string s;
int main()
{
while (cin >> n)
{
cin >> s;
tip = 0;
top = n - 1;
while (tip <= top)
{
if (s[tip] < s[top])
{
cout << s[tip];
tip++;
}
else if (s[tip] > s[top])
{
cout << s[top];
top--;
}
else
{
int ti = tip, to = top, f = 0;
while (ti <= to)
{
ti++;
to--;
if (s[ti] < s[to])
{
f = 0;
break;
}
else if (s[ti] > s[to])
{
f = 1;
break;
}
}
if (f)
{
cout << s[top];
top--;
}
else
{
cout << s[tip];
tip++;
}
}
}
cout << endl;
}
return 0;
}
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