二分简单题

二分·二分查找

题目传送：hihoCoder - 1128 - 二分·二分查找

解法一：先排个序，然后在进行二分查找，复杂度为O（n*logn） 。

AC代码：

#include <map>#include <set>#include <cmath>#include <deque>#include <queue>#include <stack>#include <cstdio>#include <cctype>#include <string>#include <vector>#include <cstdlib>#include <cstring>#include <iostream>#include <algorithm>#define LL long long#define INF 0x7fffffffusing namespace std;int n, k;int a[1000005];int b_search(int x) {    int L = 0, R = n - 1;    while(L <= R) {        int mid = (L + R) >> 1;        if(a[mid] == x) return mid + 1;        else if(a[mid] > x) R = mid - 1;        else if(a[mid] < x) L = mid + 1;    }    return -1;}int main() {    while(scanf("%d %d", &n, &k) != EOF) {        for(int i = 0; i < n; i ++) {            scanf("%d", &a[i]);        }        sort(a, a + n);        int ans = b_search(k);        printf("%d\n", ans);    }    return 0;}

解法二：直接找是否存在k，并且有多少比k小的数，复杂度O（n）。

AC代码：

#include <map>#include <set>#include <cmath>#include <deque>#include <queue>#include <stack>#include <cstdio>#include <cctype>#include <string>#include <vector>#include <cstdlib>#include <cstring>#include <iostream>#include <algorithm>#define LL long long#define INF 0x7fffffffusing namespace std;int n, k, a;int main() {    scanf("%d %d", &n, &k);    int ans = 0;    int flag = 0;    for(int i = 0; i < n; i ++) {        scanf("%d", &a);        if(a < k) ans ++;        if(a == k) flag = 1;    }    if(flag == 1) {        printf("%d\n", ans + 1);    }    else printf("-1\n");    return 0;}

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二分·二分查找之k小数

题目传送：hihoCoder - 1133 - 二分·二分查找之k小数

解法一：直接先排个序，然后在找第k小数，复杂度为O（n*logn）。

AC代码：

#include <map>#include <set>#include <cmath>#include <deque>#include <queue>#include <stack>#include <cstdio>#include <cctype>#include <string>#include <vector>#include <cstdlib>#include <cstring>#include <iostream>#include <algorithm>#define LL long long#define INF 0x7fffffffusing namespace std;int n, k;int a[1000005];int main() {    scanf("%d %d", &n, &k);    for(int i = 1; i <= n; i ++) {        scanf("%d", &a[i]);    }    sort(a + 1, a + n + 1);    if(k > n || k < 1) printf("-1\n");    else printf("%d\n", a[k]);    return 0;}

解法二：手写一个快排，其中排除掉不需要排序的区间，这样理论上可以达到O（n）的复杂度

AC代码：

#include <map>#include <set>#include <cmath>#include <deque>#include <queue>#include <stack>#include <cstdio>#include <cctype>#include <string>#include <vector>#include <cstdlib>#include <cstring>#include <iostream>#include <algorithm>#define LL long long#define INF 0x7fffffffusing namespace std;int a[1000005];int n, k;void Qsort(int *a, int low, int high) {    if(low > k) return;    if(high < k) return;    if(low >= high) return;    int first = low;    int last = high;    int key = a[first];    while(first < last) {        while(first < last && a[last] >= key) last --;        a[first] = a[last];        while(first < last && a[first] <= key) first ++;        a[last] = a[first];    }    a[first] = key;    Qsort(a, low, first - 1);    Qsort(a, first + 1, high);}int main() {    scanf("%d %d", &n, &k);    for(int i = 1; i <= n; i ++) {        scanf("%d", &a[i]);    }    Qsort(a, 1, n);    printf("%d\n", a[k]);    return 0;}

解法三：nth_element函数，用于排序无序数组里的第n个数，位于<algorithm>头文件里

AC代码：

#include <map>#include <set>#include <cmath>#include <deque>#include <queue>#include <stack>#include <cstdio>#include <cctype>#include <string>#include <vector>#include <cstdlib>#include <cstring>#include <iostream>#include <algorithm>#define LL long long#define INF 0x7fffffffusing namespace std;int a[1000005];int main() {    int n, k;    scanf("%d %d", &n, &k);    for(int i = 1; i <= n; i ++) {        scanf("%d", &a[i]);    }    nth_element(a + 1, a + k, a + n + 1);    printf("%d\n", a[k]);    return 0;}

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