- 卡片
#include
using namespace std;
const int maxn = 2021;
int car[10];
int get(int x) {
return x % 10;
}
int main(){
for (int i = 0; i < 10; i++) {
car[i] = maxn;
}
car[0]--;
for (int k = 1;; k++) {
int i = k;
while (i) {
int x=get(i);
car[x]--;
if (car[x] == -1) {
cout << k-1 << endl;//因为这个数字已经不能拼了
}
i /= 10;
}
}
return 0;
}
- 直线
#include
#include
using namespace std;
set
double ins = 1e-5;
int main() {
double x1, y1, x2, y2;
for (x1 = 0; x1 < 20; x1++) {
for (y1 = 0; y1 < 21; y1++) {
for (x2 = 0; x2 < 20; x2++) {
for (y2 = 0; y2 < 21; y2++) {
if (x1 != x2 && y1 != y2) {
double k = (y2 - y1) / (x2 - x1);
//double b = y2 - k * x2;如果按照第一种写法，
//那么y2是int型的，int减double会精度缺失，
//像第二种解法，同分了就比较高精度。
double b = (y2 * (x2 - x1) - (y2 - y1) * x2) / (x2 - x1);
pair
newline.first = k;
newline.second = b;
line.insert(newline);
}
}
}
}
}
printf("%d", line.size() + 20 + 21);
return 0;
}
- 货物摆放
求这个数的约数用约数排列组合
因为只有是约数才有可能乘机是这个数
#include
using namespace std;
typedef long long LL;
LL n = 2021041820210418;
LL mun[101000], cnt=0;
int main() {
for (LL i = 1; i <= n/i; i++) {
if (n % i == 0) {
mun[++cnt] = i;
if (i * i != n) {
mun[++cnt] = n / i;
}
}
}
cout << cnt << endl;
int ans = 0;
for (int i = 1; i <= cnt; i++) {
for (int j = 1; j <= cnt; j++) {
if (n % (mun[i] * mun[j]) == 0) {
ans++;
}
}
}
cout << ans << endl;
return 0;
}
- 路径
#include
using namespace std;
int mp[2022][2022];
const int inf = 0x3f3f3f3f;
int d[2022];//大一点
bool vis[2022];
int gcd(int a, int b) {
return b ? gcd(b, a % b) : a;
}
int main() {
for (int i = 1; i <= 2021; i++) {
for (int j = 1; j <= 2021; j++) {
if (i == j) {
mp[i][j] = 0;
}
else if (abs(i - j) > 21) {
mp[i][j] = mp[j][i] = inf;
}
else {
mp[i][j] = mp[j][i] = i * j / gcd(i, j);
}
}
}
//floyd
//dijkstra
memset(vis, false, sizeof(vis));
memset(d, 0x3f3f3f3f, sizeof(d));
d[1] = 0;
for (int i = 1; i <= 2021; i++) {
int x = 0;
for (int j = 1; j <= 2021; j++) {
if (!vis[j]&&d[j] x = j; } } vis[x] = true; for (int j = max(1, x - 21); j <= min(2021, x + 21); j++) {//剪枝只有这些边有值才会有边 d[j] = min(d[j], d[x] + mp[x][j]); } } //cout << mp[1][2021] << endl; cout << d[2021] << endl; return 0; } #include using namespace std; int main(){ long long time,s; int h, m; cin >> time; s = time / 1000; h = s / 3600 % 24; s = s % 3600; m = s / 60 ; s = s % 60; printf("%02d:%02d:%02ld", h, m, s); return 0; } #include using namespace std; const int N = 110,M=100010; int a[N],m; bool dp[N][M]; int main() { int n; cin >> n; for (int i = 1; i <= n; i++) { cin >> a[i]; m += a[i]; } dp[0][0] = true; for (int i = 1; i <= n; i++) { for (int j = 0; j <= m; j++) { dp[i][j] = dp[i - 1][j] || dp[i - 1][j + a[i]] || dp[i - 1][abs(j - a[i])]; } }//这是分成三种情况第一个是没放的 //第二个放在右面 //第三个放在左边所以要减去 //每个都三种情况 //然后一次加入每个砝码就就可以得到结果类似广度优先搜索 int ans = 0; for (int i = 1; i <= m; i++) { if (dp[n][i]) { ans++; } } cout << ans << endl; return 0; } #include using namespace std; int a[2005][2005]; int main() { //二项式问题 //二项式定理，对于C(3，n)，当n等于2000时，C(3，2000)＞1e9 //因此只需要算到第2000行就好了，剩下的再算C(1，n)和C(2，n)就好了。 int n; cin >> n; memset(a, 0, sizeof(a)); a[0][0]=1; for (int i = 1; i <= 2005; i++) { for (int j = 1; j <= i; j++) { a[i][j] = a[i - 1][j] + a[i - 1][j - 1]; if (a[i][j] == n) { cout << i * (i - 1) / 2 + j << endl; return 0; } } } int a; a = sqrt(n * 2) + 1;//杨辉三角和二项式的关系 if (a * (a - 1) / 2 == n) {//c(2 ,n)c是排列a是组合 cout << a * (a - 1) / 2 + 3 << endl; } else { cout << n * (n - 1) / 2 + 2 << endl; } return 0; } #include #include using namespace std; int a[100010]; bool cmp(int a, int b) { return a > b; } int main() { int n, m; cin >> n >> m; for (int i = 1; i <= n; i++)a[i] = i; while (m--) { int p, q; cin >> p >> q; if (p == 0) { //sort(a + 1, a + q + 1, greater sort(a + 1, a + q + 1, cmp); }else{ sort(a + q, a + 1 + n); } } for (int i = 1; i <= n; i++) { cout << a[i] << " "; } cout << endl; return 0; } 错误 思路在右括号多时从右边开始反过来 #include #include using namespace std; int ans=0; int x = 0;//第几个符号 int lm=0, rm=0,mun,vis; string s; void bfs(int x,int l,int r) { if (x == s.size()) { if (l == r)ans++; cout << l << " " << r< return; } char c=s[x]; if (c == '(') { l++; } else { r++; } cout <<" "<< l << " " << r << endl; for (int i = abs(l - r); i >= 0; i--) { if (vis) { if (i)vis -= i; if (mun) { bfs(x + 1, r + i, l); } else { bfs(x + 1, l, r + i); } if (i)vis += i; } else if (i == 0) { bfs(x + 1, l, r); } } if (c == '(') { l--; } else { r--; } return; } int main() { cin >> s; for (int i = 0; i < s.size(); i++) { if (s[i] == '(')lm++; if (s[i] == ')')rm++; } mun = rm - lm; vis = abs(rm - lm); cout << mun << endl; bfs(0,0,0); cout << ans< return 0; }
