本文主要是介绍算法竞赛入门经典(第二版)-刘汝佳-第十章 数学概念与方法 习题(12/51),希望对大家解决编程问题提供一定的参考价值,需要的开发者们随着小编来一起学习吧!
文章目录
- 说明
- 习题
- 习10-1
- 习10-2
- 习10-3
- 习10-4
- 习10-5
- 习10-6
- 习10-7
- 习10-8
- 习10-9
- 习10-10
- 习10-11
- 习10-12
- 习10-13
- 习10-14
- 习10-15
- 习10-16
- 习10-17
- 习10-18
- 习10-19
- 习10-20
- 习10-21
- 习10-22
- 习10-23
- 习10-24
- 习10-25
- 习10-26
- 习10-27
- 习10-28
- 习10-29
- 习10-30
- 习10-31
- 习10-32
- 习10-33
- 习10-34
- 习10-35
- 习10-36
- 习10-37
- 习10-38
- 习10-39
- 习10-40
- 习10-41
- 习10-42
- 习10-43
- 习10-44
- 习10-45
- 习10-46
- 习10-47
- 习10-48
- 习10-49
- 习10-50
- 习10-51
说明
本文是我对第十章51道习题的练习总结,建议配合紫书——《算法竞赛入门经典(第2版)》阅读本文。
先贴代码,文字性题解回头再补充。
习题
习10-1
题意
思路
代码
#include <cstdio>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <vector>
using namespace std;#define FOR1(i, a, b) for (int i = (a); i <= (int)(b); i++)
#define FOR2(i, a, b) for (int i = (a); i >= (int)(b); i--)const int MAX = 100;int A[10][10];int main() {
#ifdef CODE_LIANGfreopen("datain.txt", "r", stdin);freopen("dataout.txt", "w", stdout);
#endifint T;cin >> T;FOR1(t, 1, T) {FOR1(i, 0, 8) {if (i & 1) continue;FOR1(j, 0, i) {if (j & 1) continue;cin >> A[i][j];}}FOR1(i, 0, 6) {if (i & 1) continue;FOR1(j, 0, i) {if (j & 1) continue;A[i + 2][j + 1] = (A[i][j] - A[i + 2][j] - A[i + 2][j + 2]) / 2;A[i + 1][j] = A[i + 2][j] + A[i + 2][j + 1];A[i + 1][j + 1] = A[i + 2][j + 1] + A[i + 2][j + 2];}}FOR1(i, 0, 8) {FOR1(j, 0, i-1) {printf("%d ", A[i][j]);}printf("%d\n", A[i][i]);}}return 0;
}
习10-2
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习10-3
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习10-4
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代码
#include <cstdio>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <vector>
using namespace std;#define FOR1(i, a, b) for (int i = (a); i <= (int)(b); i++)
#define FOR2(i, a, b) for (int i = (a); i >= (int)(b); i--)const int MAX = 1299709+100000;//确保下一个素数被包含int n;
int isprime[MAX];
vector<int> prime;void process_prime() {memset(isprime, 0x3f, sizeof(isprime));isprime[1] = 0;FOR1(i, 2, MAX) {if (isprime[i]) {prime.push_back(i);for (int j = i * 2; j <= MAX; j += i) {isprime[j] = 0;}}}
}int main() {
#ifdef CODE_LIANGfreopen("datain.txt", "r", stdin);freopen("dataout.txt", "w", stdout);
#endifprocess_prime();while (cin >> n && n) {if (isprime[n]) {printf("0\n", n);continue;}int s = prime.size();FOR1(i, 0, s - 1) {if (prime[i] > n) {printf("%d\n", prime[i] - prime[i - 1]);break;}}}return 0;
}
习10-5
题意
思路
代码
#include <cstdio>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <vector>
using namespace std;#define FOR1(i, a, b) for (int i = (a); i <= (int)(b); i++)
#define FOR2(i, a, b) for (int i = (a); i >= (int)(b); i--)const int MAXN = 1120;
const int MAXP = 187; //计算得出最多有187个素数
const int MAXK = 14; //计算得出最多有187个素数int n, k, ans;
int isprime[MAXN+1];
vector<int> prime;
//int dp[MAXP + 1][MAXK + 1][MAXN + 1]; //dp[i][j][m]表示前i个素数中挑出j个素数其值为m的可能数,实际上第一维可以省略
int dp[MAXK + 1][MAXN + 1]; //dp[j][m]表示前i个素数中挑出j个素数其值为m的可能数,第一维已经省略void process_prime() {memset(isprime, 0x3f, sizeof(isprime));isprime[1] = 0;FOR1(i, 2, MAXN) {if (isprime[i]) {prime.push_back(i);for (int j = i * 2; j <= MAXN; j += i) {isprime[j] = 0;}}}
}int main() {
#ifdef CODE_LIANGfreopen("datain.txt", "r", stdin);freopen("dataout.txt", "w", stdout);
#endifprocess_prime();while (cin >> n >> k && n) {memset(dp, 0, sizeof(dp));dp[0][0] = 1;FOR1(i, 1, MAXP) {if (prime[i-1] > n) break;int pi = prime[i-1];FOR2(j, min(i, k), 1) {FOR2(m, n, pi) {dp[j][m] += dp[j - 1][m - pi];}}}printf("%d\n", dp[k][n]);}return 0;
}
习10-6
题意
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#include <cstdio>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <vector>
using namespace std;#define FOR1(i, a, b) for (int i = (a); i <= (int)(b); i++)
#define FOR2(i, a, b) for (int i = (a); i >= (int)(b); i--)const int MAX = 10000;//确保下一个素数被包含int n, p; //p表示prime中素数个数
int isprime[MAX + 1];
vector<int> prime;void process_prime() {memset(isprime, 0x3f, sizeof(isprime));isprime[1] = 0;FOR1(i, 2, MAX) {if (isprime[i]) {prime.push_back(i);for (int j = i * 2; j <= MAX; j += i) {isprime[j] = 0;}}}p = prime.size();
}int main() {
#ifdef CODE_LIANGfreopen("datain.txt", "r", stdin);freopen("dataout.txt", "w", stdout);
#endifprocess_prime();while (cin >> n && n) {int res = 0;int i = 0, j = 0;int s = prime[0];while (j < p) {if (s == n) {res++;if (j == p - 1) break;s += prime[++j];}else if (s < n) {if (j == p - 1) break;s += prime[++j];}else {if (i == p - 1) break;s -= prime[i++];if (i > j)s += prime[++j];}}printf("%d\n", res);}return 0;
}
习10-7
题意
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#include <cstdio>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <vector>
using namespace std;typedef long long LL;#define FOR1(i, a, b) for (int i = (a); i <= (int)(b); i++)
#define FOR2(i, a, b) for (int i = (a); i >= (int)(b); i--)const LL MAX = 1000000;
const LL LMAX = MAX*MAX;LL a[2];
int p; //p表示oneprime中素数个数
int isprime[MAX + 1];
vector<LL> oneprime;void process_prime() {memset(isprime, 0x3f, sizeof(isprime));isprime[1] = 0;FOR1(i, 2, MAX) {if (isprime[i]) {for (LL j = (LL)i * i; j <= LMAX; j *= i)oneprime.push_back(j);for (int j = i * 2; j <= MAX; j += i) {isprime[j] = 0;}}}sort(oneprime.begin(), oneprime.end());p = oneprime.size();
}int main() {
#ifdef CODE_LIANGfreopen("datain.txt", "r", stdin);freopen("dataout.txt", "w", stdout);
#endifprocess_prime();int T;cin >> T;FOR1(t, 1, T) {cin >> a[0] >> a[1];a[1]++;int res[2];FOR1(j, 0, 1) {FOR1(i, 0, p - 1) {if (oneprime[i] >= a[j]) {res[j] = i;break;}}}printf("%d\n", res[1] - res[0]);}return 0;
}
习10-8
题意
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#include <cstdio>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <vector>
#include <map>
using namespace std;typedef long long LL;#define FOR1(i, a, b) for (int i = (a); i <= (int)(b); i++)
#define FOR2(i, a, b) for (int i = (a); i >= (int)(b); i--)const LL MAX = 500000;
const LL LMAX = MAX*MAX;int n, n0;
map<LL, int> oneprime;void process_prime() {FOR1(i, 2, MAX) {if (!oneprime.count(i)) {int k = 2;for (LL j = (LL)i * i; j <= LMAX; j *= i, k++)if (!oneprime.count(j)) oneprime[j] = k;}}
}int main() {
#ifdef CODE_LIANGfreopen("datain.txt", "r", stdin);freopen("dataout.txt", "w", stdout);
#endifprocess_prime();while (cin >> n && n) {if (n == (-(1<<30) - (1<<30))) {printf("31\n");continue;}n0 = abs(n);if (oneprime.count(n0)) {int ans = oneprime[n0];if (n < 0) {while (ans % 2 == 0) ans >>= 1;}printf("%d\n", ans);}elseprintf("1\n");}return 0;
}
习10-9
题意
思路
代码
#include <cstdio>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <vector>
using namespace std;#define FOR1(i, a, b) for (int i = (a); i <= (int)(b); i++)
#define FOR2(i, a, b) for (int i = (a); i >= (int)(b); i--)//const int MAXN2 = 1e9;
const int MAXN = 40000;int L, U;
int isprime[MAXN+1];
vector<int> prime;
int resn, resi;void process_prime() {memset(isprime, 0x3f, sizeof(isprime));isprime[1] = 0;FOR1(i, 2, MAXN) {if (isprime[i]) {prime.push_back(i);for (int j = i * 2; j <= MAXN; j += i) {isprime[j] = 0;}}}
}int main() {
#ifdef CODE_LIANGfreopen("datain.txt", "r", stdin);freopen("dataout.txt", "w", stdout);
#endifprocess_prime();int T;cin >> T;FOR1(t, 1, T) {cin >> L >> U;resn = L, resi = 0;FOR1(k, L, U) {int k1 = k;int ik = k1 - L;int res = 1;int k2 = sqrt(k1) + 1;int pcnt = prime.size();FOR1(j, 0, pcnt-1) {int mult = 1;int pj = prime[j];if (k1 == 1) break;while (k1 % pj == 0) {k1 /= pj;mult++;}res *= mult;}if (k1 > 1) //说明还有约数res *= 2;if (res > resi) {resn = k;resi = res;}}printf("Between %d and %d, %d has a maximum of %d divisors.\n", L, U, resn, resi);}return 0;
}
习10-10
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习10-11
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#include <cstdio>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <vector>
#include <string>
using namespace std;#define FOR1(i, a, b) for (int i = (a); i <= (int)(b); i++)
#define FOR2(i, a, b) for (int i = (a); i >= (int)(b); i--)const int MAXN = 10000;int n;
string s[3];string add(string s1, string s2) {string sum;int up = 0;int l1 = s1.size(), l2 = s2.size();FOR1(i, 0, MAXN) {if (i >= l1 && i >= l2) {if (up) sum += (char)(up + 48);break;}int a1 = (i < l1) ? s1[i]-48 : 0;int a2 = (i < l2) ? s2[i]-48 : 0;up = up + a1 + a2;sum += (char)(up % 10 + 48);up /= 10;}return sum;
}int main() {
#ifdef CODE_LIANGfreopen("datain.txt", "r", stdin);freopen("dataout.txt", "w", stdout);
#endifwhile (cin >> n) {s[1] = "1";s[2] = "2";FOR1(i, 0, n - 3) {s[0] = s[1];s[1] = s[2];s[2] = add(s[0], s[1]);}s[1] = add(s[0], s[2]);int scnt = s[1].size();FOR2(i, scnt - 1, 0)printf("%c", s[1][i]);printf("\n");}return 0;
}
习10-12
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#include <cstdio>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <vector>
#include <string>
using namespace std;#define FOR1(i, a, b) for (int i = (a); i <= (int)(b); i++)
#define FOR2(i, a, b) for (int i = (a); i >= (int)(b); i--)const int MAXN = 50000;int n;
double F[MAXN + 1];void pre_process() {F[1] = 1;F[2] = 0.5;FOR1(i, 2, MAXN - 1)F[i + 1] = F[i] * (2 * i - 1) / 2 / i;
}int main() {
#ifdef CODE_LIANGfreopen("datain.txt", "r", stdin);freopen("dataout.txt", "w", stdout);
#endifpre_process();int T;cin >> T;FOR1(t, 1, T) {cin >> n;printf("%.4lf\n", 1 - F[n / 2]);}return 0;
}
习10-13
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习10-14
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习10-15
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习10-16
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#include <cstdio>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <vector>
#include <string>
using namespace std;#define FOR1(i, a, b) for (int i = (a); i <= (int)(b); i++)
#define FOR2(i, a, b) for (int i = (a); i >= (int)(b); i--)const int MAXN = 1000;int n;
string S[1001];string add(string s1, string s2) {string sum;int up = 0;int l1 = s1.size(), l2 = s2.size();FOR1(i, 0, MAXN) {if (i >= l1 && i >= l2) {if (up) sum += (char)(up + 48);break;}int a1 = (i < l1) ? s1[i] - 48 : 0;int a2 = (i < l2) ? s2[i] - 48 : 0;up = up + a1 + a2;sum += (char)(up % 10 + 48);up /= 10;}return sum;
}void pre_process() {S[1] = "0";S[2] = "1";S[3] = "1";FOR1(i, 4, MAXN)S[i] = add(S[i-2], add(S[i-1], S[i-2]));
}int main() {
#ifdef CODE_LIANGfreopen("datain.txt", "r", stdin);freopen("dataout.txt", "w", stdout);
#endifpre_process();while (cin >> n) {int scnt = S[n].size();FOR2(i, scnt - 1, 0)printf("%c", S[n][i]);printf("\n");}return 0;
}
习10-17
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#include <iostream>
#include <cstdio>
#include <cmath>
using namespace std;const int N = 1000001;bool isPrime[N+1];
bool isSemi[N+1];
int countSemi[N/4+1];void initPrime()
{ int i;for (i = 0; i <= N; i ++)isPrime[i] = true;isPrime[0] = isPrime[1] = false;for (i = 5; i <= N; i += 4) {if (isPrime[i]) {if (i > (int)sqrt((double)N)) continue;for (int j = i*i; j <= N; j += 4*i)isPrime[j] = false;}}
}void initSemi()
{int i, j;for (i = 1; i <= N; i += 4) {isSemi[i] = false;countSemi[i/4] = 0;}for (i = 5; i <= N; i += 4) {if (i > (int)sqrt((double)N)) break;for (j = i; j <= N; j += 4) {if (i * j > N) break;if (isPrime[i] && isPrime[j])isSemi[i * j] = true;}}for (i = 5; i <= N; i += 4) {countSemi[i/4] = countSemi[i/4-1];if (isSemi[i])countSemi[i/4] ++;}
}int main(void)
{int n;initPrime();initSemi();while (cin >> n && n) {printf("%d %d\n", n, countSemi[n/4]);}return 0;
}
习10-18
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习10-19
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习10-20
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#include <cstdio>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <vector>
#include <string>
using namespace std;#define FOR1(i, a, b) for (int i = (a); i <= (int)(b); i++)
#define FOR2(i, a, b) for (int i = (a); i >= (int)(b); i--)const int MAXM = 2000;int n, m;
int u[MAXM], d[MAXM];int main() {
#ifdef CODE_LIANGfreopen("datain.txt", "r", stdin);freopen("dataout.txt", "w", stdout);
#endifwhile (cin >> n >> m) {int ans = -1, minfloor = 0x3f3f3f3f;FOR1(i, 0, m - 1) {cin >> u[i] >> d[i];int floor = (n*u[i]) - (n*u[i] - 1) / (u[i] + d[i]) * (u[i] + d[i]);if (floor < minfloor) {ans = i;minfloor = floor;}}printf("%d\n", minfloor);}return 0;
}
习10-21
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习10-22
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习10-23
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习10-24
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习10-25
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习10-26
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习10-27
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习10-28
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习10-29
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习10-30
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习10-31
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习10-32
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习10-33
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习10-34
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习10-35
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习10-36
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习10-37
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习10-38
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习10-39
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习10-40
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习10-41
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习10-42
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习10-43
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习10-44
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习10-45
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习10-46
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习10-47
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习10-48
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习10-49
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习10-50
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习10-51
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