Codeforces Round #315 (Div. 2)

题目传送：Codeforces Round #315 (Div. 2)

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A. Music

题意较难懂，不过只要推公式就好了

注意到S+(q - 1) * t = q * t;

只需要t等于S即可，即每次增加S秒，就需要重新听一次歌

AC代码：

#include <map>#include <set>#include <list>#include <cmath>#include <deque>#include <queue>#include <stack>#include <bitset>#include <cctype>#include <cstdio>#include <string>#include <vector>#include <complex>#include <cstdlib>#include <cstring>#include <fstream>#include <sstream>#include <utility>#include <iostream>#include <algorithm>#include <functional>#define LL long long#define INF 0x7fffffffusing namespace std;int main() {    int T, S, q;    scanf("%d %d %d", &T, &S, &q);    int ans = 0;    while(S < T) {        S = S * q;        ans ++;    }    cout << ans << endl;    return 0;}

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B. Inventory

水题。。

AC代码：

#include <map>#include <set>#include <list>#include <cmath>#include <deque>#include <queue>#include <stack>#include <bitset>#include <cctype>#include <cstdio>#include <string>#include <vector>#include <complex>#include <cstdlib>#include <cstring>#include <fstream>#include <sstream>#include <utility>#include <iostream>#include <algorithm>#include <functional>#define LL long long#define INF 0x7fffffffusing namespace std;int n;int vis[100005];int ans[100005];int pos[100005];int pos_cnt;int main() {    pos_cnt = 0;    scanf("%d", &n);    for(int i = 1; i <= n; i ++){        int t;        scanf("%d", &t);        if(t <= n && t >= 1 && vis[t] == 0) {            ans[i] = t;            vis[t] = 1;        }        else {            pos[pos_cnt ++] = i;        }    }    int p = 1;    for(int i = 0; i < pos_cnt; i ++) {        for(;p <= n; p ++) {            if(vis[p] == 0) {                ans[pos[i]] = p;                vis[p] = 1;                break;            }        }    }    for(int i = 1; i < n; i ++) {        printf("%d ", ans[i]);    }    printf("%d\n", ans[n]);    return 0;}

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C. Primes or Palindromes?

枚举大法好。。

AC代码：

#include <map>#include <set>#include <list>#include <cmath>#include <deque>#include <queue>#include <stack>#include <bitset>#include <cctype>#include <cstdio>#include <string>#include <vector>#include <complex>#include <cstdlib>#include <cstring>#include <fstream>#include <sstream>#include <utility>#include <iostream>#include <algorithm>#include <functional>#include <ctime>#define LL long long#define INF 0x7fffffffusing namespace std;const int maxn = 2000005;int pi[maxn];int vis[maxn];int hw[maxn];void init() {    pi[1] = 0;    for(int i = 2; i < maxn; i ++) {        if(!vis[i]) {            pi[i] = pi[i - 1] + 1;            for(int j = 2 * i; j < maxn; j += i) {                vis[j] = 1;            }        }        else pi[i] = pi[i-1];    }}int p, q;int search() {    for(int i = maxn - 1; i >= 0; i --) {        if((LL)pi[i] * q <= (LL)hw[i] * p) return i;    }}bool fun(int n) {    int m = 0;    int t = n;    while(t) {        m = m * 10 + t % 10;        t /= 10;    }    //cout << m << " " << n << endl;    return m == n;}int main() {    init();    hw[0] = 0;    for(int i = 1; i < maxn; i++) {        if(fun(i)) hw[i] = hw[i-1] + 1;        else hw[i] = hw[i-1];    }    scanf("%d %d", &p, &q);    int ans = search();    printf("%d\n", ans);    return 0;}

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D. Symmetric and Transitive

题意：就是去求在一个含有n个元素的集合里，满足对称性和传递性，不满足自反性的关系有多少种。

这里有一个奇怪的东西——Bell数

Bell数，表示基数为n的集合划分数目，也就是对应的等价关系个数

可以发现一个奇怪的规律：ans[n] = Bell[n +１] - Bell[n];

然后根据Bell三角形打表就可以了

AC代码：

#include <map>#include <set>#include <list>#include <cmath>#include <deque>#include <queue>#include <stack>#include <bitset>#include <cctype>#include <cstdio>#include <string>#include <vector>#include <complex>#include <cstdlib>#include <cstring>#include <fstream>#include <sstream>#include <utility>#include <iostream>#include <algorithm>#include <functional>#define LL long long#define INF 0x7fffffffusing namespace std;const int MOD = 1e9+7;LL Bell[4005][4005];int main() {    int n;    scanf("%d", &n);    Bell[0][0] = 1;    for(int i = 1; i <= n; i ++) {        Bell[i][0] = Bell[i - 1][i - 1];        for(int j = 1; j <= i; j ++) {            Bell[i][j] = (Bell[i][j - 1] + Bell[i - 1][j - 1]) % MOD;        }    }    printf("%I64d\n", Bell[n][n - 1]);    return 0;}

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