poj.org 部分答案（二）
2336
#define _CRT_SECURE_NO_WARNINGS
#include
#include
class Solution {
public:
Solution() {
std::cin >> n;
number1 = new int[n];
for (int i = 0; i < n; i++) scanf("%d", &number1[i]);
std::cin >> m;
number2 = new int[m];
for (int i = 0; i < m; i++) scanf("%d", &number2[i]);
}
int binarySearch(int arr[], int target, int left, int right) {
if (left <= right) {
int mid = (left + right) / 2;
if (arr[mid] == target) {
return 1;
}
else if (arr[mid] < target) {
return binarySearch(arr, target, mid + 1, right); // 修正更新边界值
}
else {
return binarySearch(arr, target, left, mid - 1); // 修正更新边界值
}
}
return 0;
}
void solve() {
bool found = false; // 添加一个变量来标记是否找到匹配的数字对
for (int i = 0; i < m; i++) {
if (binarySearch(number1, 10000 - number2[i], 0, n - 1)) {
std::cout << "YES"; // 添加换行符
found = true; // 标记找到匹配的数字对
break; // 找到即可跳出循环
}
}
if (!found) {
std::cout << "NO"; // 如果没有找到匹配的数字对，输出 "NO"
}
}
int n, m;
int* number1, * number2;
};
int main()
{
Solution a;
a.solve();
return 0;
}
2503
#include
#include
#include
#include 
3714
#include
#include
#include
#include
using namespace std;
struct node
{
double x;
double y;
int type;
}a[200005];
int n;
vector temp;
bool cmp1(node &a,node &b)
{
return a.x> 1;
double d1 = solve(l, mid);
double d2 = solve(mid + 1, r);
double d = min(d1, d2);
int i;
for(i=l;i<=r;i++)
{
if(fabs(a[i].x-a[mid].x)>t;
while(t--)
{
int n;
cin>>n;
for(i=1;i<=n;i++)
{
cin>>a[i].x>>a[i].y;
a[i].type=1;
}
for(i=n+1;i<=2*n;i++)
{
cin>>a[i].x>>a[i].y;
a[i].type=2;
}
sort(a+1,a+2*n+1,cmp1);
double ans=solve(1,2*n);
printf("%.3lf\n",ans);
}
return 0;
}
3233
这个题目用记忆算法的话会爆
#include
using namespace std;
class Solution {
public:
Solution() {
cin >> n >> k >> m;
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
cin >> contain[i][j];
if(i == j) tmp[i][j] = 1;
else tmp[i][j] = 0;
sum[i][j] = 0;
}
}
}
void mul() {
int t[31][31];
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
t[i][j] = 0;
for (int x = 0; x < n; x++) {
t[i][j] = (t[i][j] + tmp[i][x] * contain[x][j]) % m;
}
}
}
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
tmp[i][j] = t[i][j];
sum[i][j] = (sum[i][j] + tmp[i][j]) % m;
}
}
}
void fun() {
for (int i = 0; i < k; i++) {
mul();
}
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
cout << sum[i][j] << " ";
}
cout << "\n";
}
}
private:
int n, k, m; // 进行 k 次
int contain[31][31];
int tmp[31][31];
int sum[31][31];
};
int main() {
Solution a;
a.fun();
system("pause");
return 0;
} 我们改用快速幂
#include
#include
class Solution {
private:
int b[65][65];
int x[65][65];
int n, k, m;
void mul(int a[65][65], int x[65][65]) {
int i, j, k;
int t[65][65];
memset(t, 0, sizeof(t));
for (i = 0; i < n; i++) {
for (j = 0; j < n; j++) {
for (k = 0; k < n; k++) {
t[i][j] = (t[i][j] + a[i][k] * x[k][j]) % m;
}
}
}
memcpy(a, t, sizeof(t));
}
public:
void solve() {
int i, j;
std::cin >> n >> k >> m;
for (i = 0; i < n; i++) {
for (j = 0; j < n; j++) {
std::cin >> x[i][j];
}
}
for (i = 0; i < n; i++) {
x[i + n][i] = 1;
x[i + n][i + n] = 1;
b[i][i] = 1;
b[i][i + n] = 1;
}
n <<= 1;
for (i = 0; (1 << i) <= k; i++) {
if (k & (1 << i)) {
mul(b, x);
}
mul(x, x);
}
n >>= 1;
for (i = 0; i < n; i++) {
b[i][i] = (b[i][i] - 1 + m) % m;
for (j = 0; j < n - 1; j++) {
std::cout << b[i][j] << " ";
}
std::cout << b[i][n - 1] << std::endl;
}
}
};
int main() {
Solution solution;
solution.solve();
return 0;
} #include
#include
int b[65][65], x[65][65];
int n, k, m;
void mul(int a[65][65], int x[65][65])
{
int i, j, k;
int t[65][65];
memset(t, 0, sizeof(t));
for (i = 0; i < n; i++)
for (j = 0; j < n; j++)
for (k = 0; k < n; k++)
t[i][j] = (t[i][j] + a[i][k] * x[k][j]) % m;
memcpy(a, t, sizeof(t));
}
int main()
{
int i, j;
scanf("%d%d%d", &n, &k, &m);
for (i = 0; i < n; i++)
for (j = 0; j < n; j++)
scanf("%d", &x[i][j]);
for (i = 0; i < n; i++)
{
x[i + n][i] = 1;
x[i + n][i + n] = 1;
b[i][i] = 1;
b[i][i + n] = 1;
}
n <<= 1;
for (i = 0; (1 << i) <= k; i++)
{
if (k & (1 << i))
mul(b, x);
mul(x, x);
}
n >>= 1;
for (i = 0; i < n; i++)
b[i][i] = (b[i][i] - 1 + m) % m;
for (i = 0; i < n; i++)
{
for (j = 0; j < n - 1; j++)
printf("%d ", b[i][j]);
printf("%d\n", b[i][n - 1]);
}
return 0;
}
