算法竞赛入门经典(第二版)-刘汝佳-第十章 数学概念与方法 例题（16/29）
文章目录
- 说明
- 例题
- 例10-1
- 例10-2
- 例10-3
- 例10-4
- 例10-5 （未尝试）
- 例10-6
- 例10-7
- 例10-8
- 例10-9
- 例10-10
- 例10-11
- 例10-12
- 例10-13
- 例10-14
- 例10-15
- 例10-16 （未尝试）
- 例10-17 （未尝试）
- 例10-18 （未尝试）
- 例10-19 （未尝试）
- 例10-20 （未尝试）
- 例10-21 （未尝试）
- 例10-22
- 例10-23 （未尝试）
- 例10-24
- 例10-25 （其后皆未尝试）
- 例10-26
- 例10-27
- 例10-28
- 例10-29
说明
本文是我对第十章29道例题的练习总结，建议配合紫书——《算法竞赛入门经典（第2版）》阅读本文。
先把通过的代码贴上，文字性说明以后再补。
例题
例10-1
题意
思路
代码
#include
#include
#include
#include
#include
using namespace std;
typedef unsigned long long ULL;
#define FOR1(i, a, b) for (int i = (a); i <= (b); i++)
#define FOR2(i, a, b) for (int i = (a); i >= (b); i--)
const int MAXN = 1005;
ULL a, b;
int n, n2;
//int firsti;
int F[MAXN][6*MAXN], C[MAXN * MAXN], P[MAXN];
int fast_mod(int a1, ULL b1, int p) {
if (b1 == 0) return 1;
int res = fast_mod(a1, b1 / 2, p);
res = res * res % p;
if (b1 & 1) res *= a1;
return res % p;
}
int Quick_pow_mod(int x, ULL y, int mod){
int ans = 1;
while (y){
if (y & 1) ans = (int)((ans * x) % mod);
x = (x * x) % mod;
y >>= 1;
}
return ans;
}
void pre_process()
{
FOR1(n, 2, 1000) {
F[n][0] = 0; F[n][1] = 1;
memset(C, 0, sizeof(C));
C[0 + 1*MAXN] = 1;
FOR1(i, 2, MAXN*MAXN) {
F[n][i] = (F[n][i - 1] + F[n][i - 2]) % n;
int &c = C[F[n][i - 1] + F[n][i] * MAXN];
if (c) {
P[n] = i - c;
break;
}
c = i;
}
if (n == 1000)
n = 1000;
}
}
int main() {
#ifdef CODE_LIANG
freopen("datain.txt", "r", stdin);
freopen("dataout.txt", "w", stdout);
#endif
pre_process();
int T;
scanf("%d", &T);
FOR1(t, 1, T) {
scanf("%llu %llu %d", &a, &b, &n);
int ans = 0;
if (n > 1 && a > 0) {
//int k = fast_mod(a%P[n], b, P[n]);
int k = Quick_pow_mod(a%P[n], b, P[n]);
//if (k < firsti) k += p;
ans = F[n][k];
}
printf("%d\n", ans);
}
return 0;
}
例10-2
题意
思路
代码
#include
#include
#include
#include
#include
#include
using namespace std;
typedef unsigned long long ULL;
#define FOR1(i, a, b) for (int i = (a); i <= (b); i++)
#define FOR2(i, a, b) for (int i = (a); i >= (b); i--)
const int MAXN = 101;
int T;
int X[2*MAXN];
int a0, b0;
vector B, C;
int main() {
#ifdef CODE_LIANG
freopen("datain.txt", "r", stdin);
freopen("dataout.txt", "w", stdout);
#endif
scanf("%d", &T);
FOR1(t, 1, T)
scanf("%d", &X[t]);
a0 = 0;
b0 = 0;
FOR1(a, 0, 10000) {
B.clear();
FOR1(b, 0, 10000)
B.push_back(b);
FOR1(t, 1, T - 1) {
int a2x = ((a * a) % 10001 * X[t]) % 10001;
FOR2(j, B.size() - 1, 0) {
if ((a2x + B[j] * (a + 1)) % 10001 != X[t + 1]) {
B[j] = B[B.size() - 1];
B.pop_back();
int count = B.size();
}
}
/*
//vector是连续存储空间,只提供高效的尾部删除方法pop_back() ,在中间删除的效率很低，因此该代码不行
for (vector::iterator it = B.begin(); it != B.end(); ) {
if ((ax2 + (*it) * (X[t] + 1)) % 10001 != X[t + 1])
it = B.erase(it);
else
++it;
}*/
if (B.empty()) break;
}
if (!B.empty()) {
a0 = a;
b0 = B[0];
break;
}
}
//printf("%d %d\n", a0, b0);
FOR1(t, 1, T)
printf("%d\n", (a0 * X[t] + b0) % 10001);
return 0;
}
例10-3
题意
思路
代码
#include
#include
#include
#include
#include
#include
#include
using namespace std;
typedef unsigned long long ULL;
#define FOR1(i, a, b) for (int i = (a); i <= (b); i++)
#define FOR2(i, a, b) for (int i = (a); i >= (b); i--)
int p, q, r, s;
int main() {
#ifdef CODE_LIANG
freopen("datain.txt", "r", stdin);
freopen("dataout.txt", "w", stdout);
#endif
while (cin >> p >> q >> r >> s) {
if (q > p - q) q = p - q;
if (s > r - s) s = r - s;
double res = 1;
while (q > 0 || s > 0) {
if (s == 0 || q > 0 && res < 1) {
res = res * p / q;
p--;
q--;
}
else {
res = res / r * s;
r--;
s--;
}
}
printf("%.5lf\n", res);
}
return 0;
}
例10-4
题意
思路
代码
#include
#include
#include
#include
#include
#include
#include
using namespace std;
#define FOR1(i, a, b) for (int i = (a); i <= (b); i++)
#define FOR2(i, a, b) for (int i = (a); i >= (b); i--)
const int MAXP = 60000;
int n;
//int isprime[MAXP]; //并不需要
vector yinshu;
/*
void pre_process_sushu() {
memset(isprime, 0x3f, sizeof(isprime));
FOR1(i, 2, MAXP) {
if (isprime[i]) {
for (int j = i * 2; j <= MAXP; j += i)
isprime[j] = 0;
}
}
}*/
int main() {
#ifdef CODE_LIANG
freopen("datain.txt", "r", stdin);
freopen("dataout.txt", "w", stdout);
#endif
int t = 0;
while (cin >> n && n) {
long long res = 0;
int count = 0;
int n2 = sqrt(n) + 1;
FOR1(i, 2, n2) {
if (n == 1) break;
if (n % i == 0) {
int add = 1;
while (n % i == 0) {
n /= i;
add *= i;
}
res += add;
count++;
}
}
if (n != 1) {
res += n;
count++;
}
if (count == 0) res = 2;
if (count == 1) res += 1;
printf("Case %d: %lld\n", ++t, res);
}
return 0;
}
例10-5 （未尝试）
题意
思路
代码
例10-6
题意
思路
代码
#include
#include
#include
#include
#include
#include
#include
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 = 100000;
int n, m;
vector prime, e;
int neg_count;
void process_prime(int m) {
neg_count = 0;
prime.clear();
e.clear();
int m2 = sqrt(m) + 1;
FOR1(i, 2, m2) {
int k = 0;
while (m % i == 0) {
m /= i;
k++;
}
if (k) {
prime.push_back(i);
e.push_back(-k);
neg_count++;
}
}
if (m > 1) {
prime.push_back(m);
e.push_back(-1);
neg_count++;
}
}
void add_exp(int x, int d) {
//for (int j = 0; j <= ((int)prime.size() - 1); j++) {
FOR1(j, 0, prime.size() - 1) {
int p = prime[j];
if (x < p) break;
while (x % p == 0) {
x /= p;
e[j] += d;
if (e[j] == -1 && d == -1) neg_count++;
if (e[j] == 0 && d == 1) neg_count--;
}
}
}
int main() {
#ifdef CODE_LIANG
freopen("datain.txt", "r", stdin);
freopen("dataout.txt", "w", stdout);
#endif
while (cin >> n >> m) {
process_prime(m);
vector res;
n--;
FOR1(i, 0, n) {
if (i) {
add_exp(n - i + 1, 1);
add_exp(i, -1);
}
if (neg_count == 0)
res.push_back(i + 1);
}
printf("%d\n", res.size());
if (res.size() == 0) printf("\n");
else {
FOR1(i, 0, res.size() - 1)
printf("%d%c", res[i], (i == res.size() - 1) ? '\n' : ' ');
}
}
return 0;
}
例10-7
题意
思路
代码
#include
#include
#include
#include
#include
#include
#include
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;
int phi[MAXN + 1], res[MAXN + 1];
void pre_process() {
FOR1(i, 1, MAXN)
phi[i] = i;
int prime[MAXN + 1];
memset(prime, 0x3f, sizeof(prime));
FOR1(i, 2, MAXN) {
if (prime[i]) {
for (int j = i; j <= MAXN; j += i) {
if (j > i) prime[j] = 0;
phi[j] = phi[j] /i * (i - 1);
}
}
}
res[1] = 1;
FOR1(i, 2, MAXN)
res[i] = res[i-1] + phi[i] * 2;
}
//1、预先计算好所有的n。2、求每个数的所有素因子，用欧拉公式，筛法作为程序主框架。
int main() {
#ifdef CODE_LIANG
freopen("datain.txt", "r", stdin);
freopen("dataout.txt", "w", stdout);
#endif
pre_process();
while (cin >> n && n) {
printf("%d\n", res[n]);
}
return 0;
}
例10-8
题意
思路
代码
#include
#include
#include
#include
#include
#include
#include
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--)
int k;
char s1[7][7], s2[7][7];
vector same[5];
int cnt[5], mult[5];
int main() {
#ifdef CODE_LIANG
freopen("datain.txt", "r", stdin);
freopen("dataout.txt", "w", stdout);
#endif
int T;
cin >> T;
FOR1(t, 1, T) {
scanf("%d", &k);
char st[7];
FOR1(i, 0, 5) {
scanf("%s", st);
FOR1(j, 0, 4) s1[j][i] = st[j];
}
FOR1(i, 0, 5) {
scanf("%s", st);
FOR1(j, 0, 4) s2[j][i] = st[j];
}
FOR1(j, 0, 4) {
same[j].clear();
sort(s1[j], s1[j] + 6);
sort(s2[j], s2[j] + 6);
int i1 = 0, i2 = 0;
while (i1 < 6 && i2 < 6) {
if (s1[j][i1] == s2[j][i2]) {
if (same[j].size() == 0 || same[j][same[j].size() - 1] != s1[j][i1])
same[j].push_back(s1[j][i1]);
i1++; i2++;
}
else if (s1[j][i1] < s2[j][i2])
i1++;
else
i2++;
}
cnt[j] = same[j].size();
}
mult[4] = 1;
FOR2(j, 3, 0)
mult[j] = mult[j + 1] * cnt[j + 1];
int res[5];
k--;
bool ok = true;
if (mult[0] == 0 || cnt[0] == 0) ok = false;
if (ok == true) {
FOR1(j, 0, 4) {
res[j] = k / mult[j];
k %= mult[j];
}
}
if (ok == false || res[0] >= cnt[0])
printf("NO");
else {
FOR1(j, 0, 4)
printf("%c", same[j][res[j]]);
}
printf("\n");
}
return 0;
}
例10-9
题意
思路
代码
#include
#include
#include
#include
#include
#include
#include
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--)
int n;
char s[101];
int main() {
#ifdef CODE_LIANG
freopen("datain.txt", "r", stdin);
freopen("dataout.txt", "w", stdout);
#endif
while (scanf("%s", s) != EOF) {
n = strlen(s);
int p1 = 0, p2 = 0;
FOR1(i, 0, n - 1) {
if (s[i] == '0' && s[(i + 1) % n] == '0')
p1++;
if (s[i] == '0')
p2++;
}
int res = p1 * n - p2 * p2;
if (res > 0)
printf("SHOOT\n");
else if (res < 0)
printf("ROTATE\n");
else
printf("EQUAL\n");
}
return 0;
}
例10-10
题意
思路
代码
#include
#include
#include
#include
#include
#include
#include
using namespace std;
#define FOR1(i, a, b) for (int i = (a); i <= (b); i++)
#define FOR2(i, a, b) for (int i = (a); i >= (b); i--)
int a, b, c;
int main() {
#ifdef CODE_LIANG
freopen("datain.txt", "r", stdin);
freopen("dataout.txt", "w", stdout);
#endif
while (cin >> a >> b >> c) {
double p = ((double)a * b + b * (b - 1)) / ((a + b) * (a + b - c - 1));
printf("%.5lf\n", p);
}
return 0;
}
例10-11
题意
思路
代码
#include
#include
#include
#include
#include
#include
#include
using namespace std;
#define FOR1(i, a, b) for (int i = (a); i <= (b); i++)
#define FOR2(i, a, b) for (int i = (a); i >= (b); i--)
int n, r;
double p[21];
double res[21];
double solve(int m, int m1, int k) { //q表示当前概率，m表示计算到第几步，m1表示这m个人中有m1个已经买了，k表示第k个人确定会买，k=0表示没有确定
if (m1 > r) return 0;
m++;
if (m == k) return p[m] * solve(m, m1 + 1, k);
if (m > n) {
if (m1 == r) return 1;
else return 0;
}
if (m1 == r) //已经有r人买了，后面的人不会再买
return (1 - p[m]) * solve(m, m1, k);
return p[m] * solve(m, m1 + 1, k) + (1 - p[m]) * solve(m, m1, k);
}
int main() {
#ifdef CODE_LIANG
freopen("datain.txt", "r", stdin);
freopen("dataout.txt", "w", stdout);
#endif
int t = 0;
while (cin >> n >> r && n) {
FOR1(i, 1, n)
cin >> p[i];
FOR1(i, 0, n)
res[i] = solve(0, 0, i);
printf("Case %d:\n", ++t);
FOR1(i, 1, n)
printf("%.6lf\n", res[i] / res[0]);
}
return 0;
}
例10-12
题意
思路
代码
#include
#include
#include
#include
#include
#include
#include
using namespace std;
#define FOR1(i, a, b) for (int i = (a); i <= (b); i++)
#define FOR2(i, a, b) for (int i = (a); i >= (b); i--)
const int MAXD = 1953126; //5^9 = 1953125
char s[9][5];
int e[9];
int state[9];
double dp[MAXD];
void pre_process5() {
FOR2(i, 8, 0) {
if (i == 8) e[i] = 1;
else e[i] = e[i+1] * 5;
}
}
double solve(int code) {
if (dp[code] < 1.5) //表示已访问过
return dp[code];
vector left;
FOR1(i, 0, 8)
if (state[i]) left.push_back(i);
int cnt = left.size();
if (cnt == 0) return dp[code] = 1;
dp[code] = 0;
int tot = 0;
FOR1(i, 0, cnt-1) {
int &si = state[left[i]];
FOR1(j, i + 1, cnt-1) {
int &sj = state[left[j]];
if (s[left[i]][si-1] == s[left[j]][sj-1]) {
si--; sj--;
dp[code] += solve(code - e[left[i]] - e[left[j]]);
si++; sj++;
tot++;
}
}
}
if (tot)
dp[code] = dp[code] / tot;
return dp[code];
}
int main() {
#ifdef CODE_LIANG
freopen("datain.txt", "r", stdin);
freopen("dataout.txt", "w", stdout);
#endif
pre_process5();
char s0[3];
while (scanf("%s", s0) != EOF) {
FOR1(i, 0, 8) {
state[i] = 4;
FOR1(j, 0, 3) {
if (i || j) scanf("%s", s0);
s[i][j] = s0[0];
}
}
FOR1(i, 1, MAXD)
dp[i] = 10; //表示未访问，因为概率不可能大于1
dp[0] = 1;
printf("%.6lf\n", solve(e[0] * 5 - 1));
}
return 0;
}
例10-13
题意
思路
代码
#include
#include
#include
#include
#include
#include
#include
using namespace std;
#define FOR1(i, a, b) for (int i = (a); i <= (b); i++)
#define FOR2(i, a, b) for (int i = (a); i >= (b); i--)
int n;
int f[31];
void pre_process() {
memset(f, 0, sizeof(f));
f[3] = 1;
FOR1(i, 4, 30)
f[i] = f[i - 1] * 2 + (1 << (i - 4)) - f[i - 4];
}
int main() {
#ifdef CODE_LIANG
freopen("datain.txt", "r", stdin);
freopen("dataout.txt", "w", stdout);
#endif
pre_process();
while (scanf("%d", &n) && n) {
printf("%d\n", f[n]);
}
return 0;
}
例10-14
题意
思路
代码
#include
#include
#include
#include
#include
#include
#include
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 MOD = 10056;
const int MAXN = 1000;
int n;
int C[MAXN + 1][MAXN + 1], F[MAXN + 1];
// 注意不能用乘除来推，而要用这个公式：C[i][j] = (C[i - 1][j] + C[i - 1][j - 1]) % MOD;
void pre_process() {
C[0][0] = 1;
FOR1(n, 1, MAXN) {
C[n][0] = 1;
FOR1(i, 1, n) {
C[n][i] = (C[n - 1][i] + C[n - 1][i - 1]) % MOD;
}
}
F[0] = 1;
FOR1(n, 1, MAXN) {
F[n] = 0;
FOR1(i, 1, n) {
F[n] = (F[n] + C[n][i] * F[n - i]) % MOD;
}
}
}
int main() {
#ifdef CODE_LIANG
freopen("datain.txt", "r", stdin);
freopen("dataout.txt", "w", stdout);
#endif
pre_process();
int T;
cin >> T;
FOR1(t, 1, T) {
cin >> n;
printf("Case %d: %d\n", t, F[n]);
}
return 0;
}
例10-15
题意
思路
代码
#include
#include
#include
#include
#include
#include
#include
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--)
int n, L, R;
long long solve(int k, int l, int r) {
if (l <= 0 || r <= 0 || k < l+r-1) return 0;
if (k == 1) {
if (l == 1 && r == 1) return 1;
return 0;
}
if (k == 2 && l == 1 && r == 1)
k = k;
long long res = solve(k - 1, l - 1, r) + (k-2) * solve(k - 1, l, r) + solve(k - 1, l, r - 1);
//printf("%d %d %d : %d\n", k, l, r, res);
return res;
}
int main() {
#ifdef CODE_LIANG
freopen("datain.txt", "r", stdin);
freopen("dataout.txt", "w", stdout);
#endif
int T;
cin >> T;
FOR1(t, 1, T) {
cin >> n >> L >> R;
printf("%lld\n", solve(n, L, R)); //从长到短安排，第一个参数表示已经安排好的数量，后两个参数表示左看和右看能看到的数量
}
return 0;
}
例10-16 （未尝试）
题意
思路
代码
例10-17 （未尝试）
题意
思路
代码
例10-18 （未尝试）
题意
思路
代码
例10-19 （未尝试）
题意
思路
代码
例10-20 （未尝试）
题意
思路
代码
例10-21 （未尝试）
题意
思路
代码
例10-22
题意
思路
代码
#include
#include
#include
#include
#include
#include
#include
using namespace std;
typedef vector VINT;
#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--)
int a, b;
VINT solve(int x) {
x++; //因为算法统计的是小于该数的所有数
VINT vx;
FOR1(j, 0, 9) vx.push_back(0);
int k = 1, e = 0, k10;
FOR1(i, 1, 8) {
k10 = k * 10;
if (x < k10) break;
FOR1(j, 0, 9) //算低位
vx[j] += (k-k/10) * (i-1);
FOR1(j, 1, 9) //算当前位
vx[j] += k;
k *= 10; e++;
}
VINT vt;
FOR1(j, 0, 9) vt.push_back(0);
bool first = true;
while (k) {
int t = x / k;
if (first) {
FOR1(j, 0, 9) //算低位
vx[j] += k / 10 * e * (t-1);
FOR1(j, 1, t - 1) //算当前位
vx[j] += k;
first = false;
}
else {
FOR1(j, 0, 9) //算低位
vx[j] += k / 10 * e * t;
t = t;
FOR1(j, 0, t - 1) //算当前位
vx[j] += k;
t = t;
FOR1(j, 0, 9) //算高位
vx[j] += k * vt[j] * t;
t = t;
}
vt[x / k]++;
x %= k;
k /= 10;
e--;
}
return vx;
}
int main() {
#ifdef CODE_LIANG
freopen("datain.txt", "r", stdin);
freopen("dataout.txt", "w", stdout);
#endif
while (cin >> a >> b && (a || b)) {
if (a < b) swap(a, b);
b--;
VINT va = solve(a);
VINT vb = solve(b);
FOR1(i, 0, 9)
printf("%d%c", va[i] - vb[i], i < 9 ? ' ' : '\n');
}
return 0;
}
例10-23 （未尝试）
题意
思路
代码
例10-24
题意
思路
说明：代码是汝佳写的。
我之前用另外的方法做，虽然能通过题目中测试用例，但提交后WA了。汝佳的代码极其简单，我也就没有再自己重复写，直接拿过来了。
代码
#include
int main() {
int h, w;
char s[999];
while(scanf("%d%d", &h, &w) == 2) {
int ans = 0, c = 0;
while(h--) {
scanf("%s", s);
int in = 0;
for(int i = 0; i < w; i++) {
if(s[i] == '/' || s[i] == '\\') { c++; in = !in; }
else if(in) ans++;
}
}
printf("%d\n", ans + c/2);
}
return 0;
}
例10-25 （其后皆未尝试）
题意
思路
代码
例10-26
题意
思路
代码
例10-27
题意
思路
代码
例10-28
题意
思路
代码
例10-29
题意
思路
代码