“高教社杯”第三届福建省大学生程序设计竞赛

Tag	Pro.ID	Problem Title	Ratio(AC/Submit)
	1	Solve equation	48.47%(111/229)
	2	Bin & Jing in wonderland	14.29%(4/28)
	3	Floor problem	88.67%(133/150)
	4	Digits Count	17.19%(11/64)
	5	How many tuples	0.00%(0/0)
	6	Hua Rong Dao	33.33%(22/66)
	7	Mod problem	20.69%(6/29)
	8	Mountain Number	28.57%(2/7)
	9	Star	24.38%(78/320)
	10	Min Number	32.57%(85/261)
	11	Tickets	28.00%(14/50)

1、水题

#include <stdio.h>
#include <string.h>
const int N = 105;

int t, base, n1, n2;
char num1
, num2
;

int tra(char *s) {
int ans = 0;
for (int i = 0; i < strlen(s); i ++) {
if (s[i] >= '0' && s[i] <= '9')
ans = ans * base + s[i] - '0';
else
ans = ans * base + s[i] - 'a' + 10;
}
return ans;
}

void init() {
scanf("%s%s%d", num1, num2, &base);
n1 = tra(num1); n2 = tra(num2);
}

void solve() {
printf("(%d,%d)\n", n1 / n2, n1 - n1 / n2 * n2);
}

int main() {
scanf("%d", &t);
while (t--) {
init();
solve();
}
return 0;
}

第二题:组合概率

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
#define min(a,b) (a)<(b)?(a):(b)
const int N = 55;
typedef long long ll;

double c

;
int t, n, k, r, res
, Minr, vis
;
double p
, pr, pMax;

double C(int n, int m) {
double ans = 1.0;
m = min(m, n - m);
for (int i = 0; i < m; i ++)
ans = ans * (n - i) / (i + 1);
return ans;
}
void c_table() {
for (int i = 0; i <= 52; i ++)
for (int j = 0; j <= i; j ++)
c[i][j] = C(i, j);
}

void init() {
pMax = 1.0;
memset(vis, 0, sizeof(vis));
scanf("%d%d%d", &n, &k, &r);
for (int i = 1; i <= n; i ++)
scanf("%lf", &p[i]);
for (int i = 0; i < r; i ++) {
scanf("%d", &res[i]);
pMax *= p[res[i]];
vis[res[i]] ++;
}
sort(res, res + r);
Minr = res[0];
pr = 0.0;
for (int i = 1; i < Minr; i ++)
pr += p[i];
}

void fuck(double &s) {
int kk = k;
for (int i = 0; i <= n; i ++) {
s *= c[kk][vis[i]];
kk -= vis[i];
}
}

double sb(double p, int k) {
double ans = 1.0;
for (int i = 0; i < k; i ++)
ans *= p;
return ans;
}

double solve() {
double ans = 0;
double s;
for (int i = 0; i <= k - r; i ++) {
vis[0] = k - r - i;
s = pMax * sb(pr ,k - r - i) * sb(p[Minr], i);
fuck(s);
ans += s;
vis[Minr] ++;
}
return ans;
}

int main() {
c_table();
scanf("%d", &t);
while (t--) {
init();
printf("%.6lf\n", solve());
}
return 0;
}

第三题：水题

#include <stdio.h>
#include <string.h>

int t, n, l, r;
int main() {
scanf("%d", &t);
while (t--) {
scanf("%d%d%d", &n, &l, &r);
int sum = 0;
for (int i = l; i <= r; i ++)
sum += n / i;
printf("%d\n", sum);
}
return 0;
}

第四题:应该是线段树，没出。

第五题：看都没看

第六题：搜索dfs

#include <stdio.h>
#include <string.h>

const int block[3][3][2] = {{{0, 0}, {0, 1}, {0, 0}}, {{0, 0}, {1, 0}, {0, 0}}, {{0, 0}}};
const int blockn[3] = {2, 2, 1};

int t, n, vis[5][5], save[5], ans;

bool judge(int x, int y, int i, int n) {
for (int j = 0; j < blockn[i]; j ++) {
int xx = x + block[i][j][0];
int yy = y + block[i][j][1];
if (xx < 0 || xx >= n || yy < 0 || yy >= 4 || vis[xx][yy]) {
return false;
}
}
return true;
}
void dfs(int x, int y, int n, int num) {
if (num == n * 4) {
ans ++;
return;
}
if (y == 4) dfs(x + 1, 0, n, num);
if (vis[x][y]) dfs(x, y + 1, n, num);
for (int i = 0; i < 3; i ++) {
if (judge(x, y, i, n)) {
for (int j = 0; j < blockn[i]; j ++)
vis[x + block[i][j][0]][y + block[i][j][1]] = 1;
dfs(x, y + 1, n, num + blockn[i]);
for (int j = 0; j < blockn[i]; j ++)
vis[x + block[i][j][0]][y + block[i][j][1]] = 0;
}
}
}

int solve(int n) {
int an = 0;
for (int i = 0; i < n - 1; i ++)
for (int j = 0; j < 3; j ++) {
memset(vis, 0, sizeof(vis));
vis[i][j] = vis[i + 1][j] = vis[i][j + 1] = vis[i + 1][j + 1] = 1;
ans = 0;
dfs(0, 0, n, 4);
an += ans;
}
return an;
}

int main() {
for (int i = 1; i <= 4; i ++)
save[i] = solve(i);
scanf("%d", &t);
while (t--) {
scanf("%d", &n);
printf("%d\n", save
);
}
return 0;
}

第七题；没看

第八题：应该是数位DP，没出。

第九题：暴力枚举+几何

#include <stdio.h>
#include <string.h>

typedef long long ll;
const int N = 105;

int t, n;
struct point {
ll x, y;
} p
;

void init() {
scanf("%d", &n);
for (int i = 0; i < n; i ++)
scanf("%lld%lld", &p[i].x, &p[i].y);
}

bool judge(int i, int j, int k) {
return (p[j].x - p[i].x) * (p[k].x - p[i].x) + (p[j].y - p[i].y) * (p[k].y - p[i].y) > 0;
}

int solve() {
int ans = 0;
for (int i = 0; i < n; i ++)
for (int j = i + 1; j < n; j ++)
for (int k = j + 1; k < n; k ++) {
if (judge(i, j, k) && judge(j, i, k) && judge(k, i, j))
ans ++;
}
return ans;
}

int main() {
scanf("%d", &t);
while (t--) {
init();
printf("%d\n", solve());
}
return 0;
}

第十题：贪心

#include <stdio.h>
#include <string.h>

const int N = 1005;

int t, m;
char n
;

void init() {
scanf("%s%d", n, &m);
}

void swap(char &a, char &b) {
char c = a;
a = b;
b = c;
}

void solve() {
if (m != 0) {
int len = strlen(n), Min = 10, Min_v = len, start = 1, mm = 0;
for (int i = len - 1; i > 0; i --)
if (Min > n[i] - '0' && n[i] != '0') {
Min = n[i] - '0';
Min_v = i;
}
if (Min < n[0] - '0') {
swap(n[0], n[Min_v]);
mm ++;
}
while (start != len && mm != m) {
Min = 10;
for (int i = len - 1; i > start; i --)
if (Min > n[i] - '0') {
Min = n[i] - '0';
Min_v = i;
}
if (Min < n[start] - '0') {
swap(n[start], n[Min_v]);
mm ++;
}
start ++;
}
}
printf("%s\n", n);
}

int main() {
scanf("%d", &t);
while (t--) {
init();
solve();
}
return 0;
}

第十一题：并查集+欧拉回路

#include <stdio.h>
#include <string.h>

const int N = 100005;

int t, n, m, vis
, parent
, d
, s
, sn, v
, num
;

int find(int x) {
return x == parent[x] ? x : parent[x] = find(parent[x]);
}

void init() {
scanf("%d%d", &n, &m);
sn = 0;
for (int i = 0; i <= n; i ++)
parent[i] = i;
memset(vis, 0, sizeof(vis));
memset(d, 0, sizeof(d));
memset(v, 0, sizeof(v));
memset(num, 0, sizeof(num));
int a, b;
while (m--) {
scanf("%d%d", &a, &b);
vis[a] = 1; vis[b] = 1;
d[a] ++; d[b] ++;
int pa = find(a), pb = find(b);
if (pa != pb)
parent[pa] = pb;
}
}

int solve() {
int ans = 0;
for (int i = 1; i <= n; i ++) {
if (vis[i]) {
int pi = find(i);
if (!v[pi]) {
v[pi] = 1;
s[sn ++] = pi;
if (d[i] % 2) num[pi] ++;
}
else {
if (d[i] % 2)
num[pi] ++;
}
}
}
for (int i = 0; i < sn; i ++) {
ans += num[s[i]] / 2 - 1;
}
ans += sn - 1;
return ans;
}

int main() {
scanf("%d", &t);
while (t--) {
init();
printf("%d\n", solve());
}
return 0;
}