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目录
P2249 【深基13.例1】查找 - 洛谷 (luogu.com.cn)
P1102 A-B 数对 - 洛谷 (luogu.com.cn)
P1678 烦恼的高考志愿 - 洛谷 (luogu.com.cn)
P2440 木材加工 - 洛谷 (luogu.com.cn)
P2249 【深基13.例1】查找 - 洛谷 (luogu.com.cn)
比原来的二分多一步，即寻找第一次出现的位置，所以while循环中相等的情况需要额外判断是不是第一次。需要注意应该写while (left <= right)
#include<iostream>
#include<vector>
using namespace std;
int main() {int n, m;int t;cin >> n >> m;vector<int> a(n + 1);for (int i = 1; i <= n; i++) cin >> a[i];for (int i = 0; i < m; i++) {cin >> t;int res = -1;int left = 1;int right = n;while (left <= right) {int mid = left + ((right - left) >> 1);if (a[mid] > t) {right = mid - 1;}else if (a[mid] < t) {left = mid + 1;}else if (a[mid] == t) {res = mid;right = mid-1 ;}}cout << res << " ";}return 0;
}
P1102 A-B 数对 - 洛谷 (luogu.com.cn)
a-b=c转化为a=b+c,找到所有的a使b+c=a
找a的时候用二分，可以拿92分（样例3TLE力）
#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;
int main() {int n, c;scanf("%d %d", &n, &c);vector<int> a(n);for (int i = 0; i < n; i++)cin >> a[i];sort(a.begin(), a.end());int res = 0;//a-b=c转化为a=b+cfor (int i = 0; i < n; i++) {int b = a[i];//找到所有的a使b+c=aint left = 0;int right = n - 1;int t = -1;while (left <= right) {int mid = left + ((right - left) >> 1);if (a[mid] > (b + c)) right = mid - 1;else if (a[mid] < (b + c)) left = mid + 1;else {t = mid;right = mid - 1;}}if (t != -1) {while (t < n && (a[t] == b + c)) {res++;t++;}}}printf("%d\n", res);return 0;
}介绍两个非常好用的函数，方法和上面一样，详见注释
#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;
int main() {int n, c;scanf("%d %d", &n, &c);vector<int> a(n);for (int i = 0; i < n; i++)cin >> a[i];sort(a.begin(), a.end());int res = 0;for (int i = 0; i < n; i++) {res += upper_bound(a.begin(), a.end(), a[i] + c) - lower_bound(a.begin(), a.end(), a[i] + c);//upper_bound：找到最后一个与a[i] + c相等的数//lower_bound:找到第一个与a[i] + c相等的数}printf("%d\n", res);return 0;
}可惜还是过不去
原来是要开long long。。。。。。
#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;
#define int long long
signed main() {int n, c;scanf("%lld %lld", &n, &c);vector<int> a(n);for (int i = 0; i < n; i++)cin >> a[i];sort(a.begin(), a.end());int res = 0;for (int i = 0; i < n; i++) {res += upper_bound(a.begin(), a.end(), a[i] + c) - lower_bound(a.begin(), a.end(), a[i] + c);//upper_bound：找到最后一个与a[i] + c相等的数//lower_bound:找到第一个与a[i] + c相等的数}printf("%lld\n", res);return 0;
}
欧耶，过了过了过了！
P1678 烦恼的高考志愿 - 洛谷 (luogu.com.cn)
给学校排序，根据学生的分数找学校（二分），直到找到小于等于学生分数的学校
#include<iostream>
#include<algorithm>
#include<vector>
#include<cmath>
using namespace std;
#define int long long
signed main() {int m, n;cin >> m >> n;int res = 0;vector<int> sch(m);vector<int> stu(n);for (int i = 0; i < m; i++)cin >> sch[i];for (int i = 0; i < n; i++)cin >> stu[i];sort(sch.begin(), sch.end());for (int i = 0; i < n; i++) {int left = 0;int right = m - 1;while (left <= right) {int mid = left + ((right - left) >> 1);if (sch[mid] > stu[i])right = mid - 1;else if (sch[mid] < stu[i]) {if (mid + 1 == m) {left = mid;break;}else {if (sch[mid + 1] > stu[i]) {left = mid;break;}else if (sch[mid + 1] == stu[i]) {left = mid + 1;break;}else left = mid + 1;}}else {left = mid;break;}}if (left != (m - 1)) {int x = abs(sch[left] - stu[i]);int y = abs(sch[left + 1] - stu[i]);if (x > y)res += y;else res += x;}else res += abs(sch[left] - stu[i]);}cout << res << endl;return 0;
}二分的时候看起来很麻烦，虽然还是AC了
P2440 木材加工 - 洛谷 (luogu.com.cn)
对切出来的最大值进行二分
#include<iostream>
#include<vector>
using namespace std;
#define int long long
signed main() {int n, k;cin >> n >> k;vector<int> a(n);int sum = 0;int f = 0;for (int i = 0; i < n; i++) {cin >> a[i];if (a[i] == 1)f = 1;sum += a[i];}if (sum / k == 0) {cout << 0 << endl;return 0;}else {if (f == 1) {cout << 1 << endl;return 0;}}int left = 0;int right = sum / k;int res = 0;while (left <= right) {int mid = left + ((right - left) >> 1);int count = 0;for (int i = 0; i < n; i++) {count += a[i] / mid;}if (count >= k) {res = mid;left = mid + 1;}else if (count<k) {right = mid - 1;}}cout << res << endl;return 0;
}
其实如果没有那两个return 0的语句会RE两个，作者也不知道怎么回事。。。(可能是数据比较恶心)
