Codeforces Round #246 (Div. 2)

题目链接：Codeforces Round #246 (Div. 2)

A：直接找满足的人数，然后整除3就是答案

B：开一个vis数组记录每一个衣服的主场和客场出现次数。然后输出的时候主场数量加上反复的，客场数量减掉反复的

C：原来是YY乱搞的。原来是哥德巴赫猜想，一个合数能够表示为3个质数相加，然后就先打个素数表，然后从最小的数字一个个模拟往前放就可以。放的时候走的步数直接拆成都是质数就可以

D：KMP算法，利用next数组性质求前缀和后缀匹配，然后在利用累加求和求出每一个串相应的出现次数

代码：

A：

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;

int n, k, num, i;

int main() {
    scanf("%d%d", &n, &k);
    int ans = 0;
    for (i = 0; i < n; i++) {
        scanf("%d", &num);
        if (5 - num >= k) ans++;
    }
    printf("%d\n", ans / 3);
    return 0;
}

B：

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

const int N = 100005;
int vis[N][2];
int n, x[N], y[N], i;

int main() {
    scanf("%d", &n);
    for (i = 0; i < n; i++) {
        scanf("%d%d", &x[i], &y[i]);
        vis[x[i]][0]++;
        vis[y[i]][1]++;
    }
    for (i = 0; i < n; i++) {
        printf("%d %d\n", (n - 1) + vis[y[i]][0], (n - 1) - vis[y[i]][0]);
    }
    return 0;
}

C：

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
const int N = 100005;

int pri[N], ans[5 * N][2], ansn = 0;

void init() {
    int vis[N];
    memset(vis, 0, sizeof(vis));
    for (int i = 2; i < N; i++) {
        if (vis[i]) continue;
        pri[i] = 1;
        for (int j = i; j < N; j += i)
            vis[j] = 1;
    }
}

int n, num[N], v[N], i, snum[N];

void swap(int &a, int &b) {
    a ^= b;
    b ^= a;
    a ^= b;
}

int main() {
    init();
    scanf("%d", &n);
    for (i = 0; i < n; i++) {
        scanf("%d", &num[i]);
        snum[i] = num[i];
        v[num[i]] = i;
    }
    sort(snum, snum + n);
    i = 0;
    while (i < n) {
        while (v[snum[i]] != i) {
            for (int j = i; ;j++) {
                if (pri[v[snum[i]] - j + 1]) {
                    ans[ansn][0] = j + 1;
                    ans[ansn][1] = v[snum[i]] + 1;
                    ansn++;
                    int t = v[snum[i]];
                    v[snum[i]] = j;
                    v[num[j]] = t;
                    swap(num[j], num[t]);
                    break;
                }
            }
        }
        i++;
    }
    printf("%d\n", ansn);
    for (i = 0; i < ansn; i++)
        printf("%d %d\n", ans[i][0], ans[i][1]);
    return 0;
}

D：

#include <stdio.h>
#include <string.h>
#define INF 0x3f3f3f3f
const int N = 100005;

char s[N];
int next[N], n, ans[N], ansn = 0;

void get_next(char *seq, int m) {
    next[1] = 0;
    int j = next[1];
    for (int i = 2; i <= m; i++) {
        while (j && seq[i] != seq[j + 1]) j = next[j];
        if (seq[i] == seq[j + 1]) j++;
        next[i] = j;
    }
}

int vis[N];

int main() {
        int i = 0;
        scanf("%s", s + 1);
        n = strlen(s + 1);
        get_next(s, n);
        int t = next[n];
        while (t) {
            ans[ansn++] = t;
            t = next[t];
        }
        for (i = n; i > 0; i--)
            vis[next[i]]++;
        for (i = n; i > 0; i--)
            vis[next[i]] += vis[i];
        printf("%d\n", ansn + 1);
        for (i = ansn - 1; i >= 0; i--)
            printf("%d %d\n", ans[i], vis[ans[i]] + 1);
        printf("%d %d\n", n, vis[n] + 1);
    return 0;
}