大一寒假集训 二分入门

A - Cable master HDU - 1551 

题目大意：求最大长度，使得每段铁丝截得的段数等于给出的k值。

#include<algorithm>
#include <iostream>
#include<cstdio>
#include<cmath>
#include<cmath>
using namespace std;
const int maxn = 100000 + 15;
double arr[maxn];
int n, k;
bool solve(double mid) {
    long long int ans = 0;
    for (int i = 0;i < n;i++) {
        ans =ans+(int)(arr[i] / mid);
    }
    if (ans >= k)return true;
    else return false;
}
int main() {
    while (scanf("%d %d", &n, &k), n + k) {
        double max = 0;
        for (int i = 0;i < n;i++) {
            scanf("%lf", &arr[i]);
            if (arr[i] > max)max = arr[i];
        }
        double left = 0, right = max, mid;
        while (right-left>1e-5) {
            mid = (right + left) / 2;
            if (solve(mid)) { left = mid; }
            else right = mid;
        }
        printf("%.2f\n", left);//注意是left，因为mid不一定满足
    }

    return 0;
}

B - Dating with girls(1) HDU - 2578

题目大意：求给出数列中任意选两数之和为k的对数。前后可以交换。

#include<algorithm>
#include <iostream>
#include<cstdio>
#include<cmath>
#include<cmath>
using namespace std;
const int maxn = 100000 + 15;
int n, k;
int arr[maxn];
int main() {
    int t;
    scanf("%d", &t);
    while (t--) {
        scanf("%d %d", &n, &k);
        for (int i = 0;i < n;i++) {
            scanf("%d", &arr[i]);
        }
        sort(arr, arr + n);
        int num = unique(arr, arr + n) - arr;//去重
        int ans = 0;
        for (int i = 0;i < num;i++) {
            int need = k - arr[i];
            if (2 * arr[i] == k) { ans++;break; }
            else if (2 * arr[i] > k) { break; }
            int left = i, right = num - 1, mid;
            while (right >= left) {
                mid = (right + left) / 2;
                if (arr[mid] < need)left = mid + 1;
                else if(arr[mid] > need)right = mid - 1;
                else if (arr[mid] == need) { ans+=2;break; }
            }
        }
        printf("%d\n", ans);
    }
    return 0;
}

C - Pie POJ - 3122  简单的二分，π是坑点

题目大意：给出蛋糕半径，分蛋糕，且每人份只能取自一个蛋糕。

#include<algorithm>
#include <iostream>
#include<cstdio>
#include<cmath>
#include<cmath>
#define pi acos(-1)
using namespace std;
const int maxn = 100000 + 15;
int n, f;
double arr[maxn];
bool solve(double mid) {
     int ans = 0;
    for (int i = 0;i < n;i++) {
        ans +=(int)(arr[i] / mid);
    }
    if (ans >= f + 1)return true;
    else return false;
}
int main() {
    int t;
    scanf("%d", &t);
    while (t--) {
        double left = 0, right=0, mid;
        scanf("%d %d", &n, &f);
        for (int i = 0;i < n;i++) {
            scanf("%lf", &arr[i]);
            arr[i] = arr[i] * arr[i] * pi;
            if (arr[i] > right)right =arr[i];
        }
        while (right - left > 1e-5) {
            mid = (right + left) / 2;
            if (solve(mid))left = mid;
            else right = mid;
        }
        printf("%.4f\n", left);
        
    }
    return 0;
}

D - Can you solve this equation? HDU - 2199 

#include <iostream>
#include <map>
#include <string>
#include<algorithm>
#include<vector>
#include<cmath>
#define max 100010
using namespace std;
int a[max];
double judge(double x){
    return 8 * x * x * x * x + 7 * x * x * x + 2 * x * x + 3 * x + 6 ;
}
int main()
{
    int t;
    scanf("%d", &t);
    while (t--) {
        long long int y;
        scanf("%d", &y);
        if (y < 0)y = -y;
        double left = 0, right = 100;
        int flag = 0;
        double mid;
        while (right >left) {
             mid = (left + right) / 2;
            if (fabs(judge(mid) - y) < 1e-5) {
                flag = 1;break;
            }
            if (judge(mid) - y > 0)right = mid;
            else left = mid;
        }
        if (flag == 0)printf("No solution!\n");
        else printf("%.4f\n", mid);
    }
}

E - Rikka with Mutex HDU - 6261   二分，strlen放在for循环中会超时

题目大意：给定一串字符串由V和P组成，一个人的初始生命值为0，遇V+1，遇P-1，同伴中可以分享生命值，在保证生命值不为负的情况下，求最少需要多少人，能让一个人走到终点。

#include <iostream>
#include <cstdio>
#include<cstring>
using namespace std;
const int maxn = 1e5 + 10;
char str[maxn];
bool solve(int mid) {
    int life = 0, step = 0;
    int len = strlen(str);
    for (int i = 0;i < len;i++) {
        if (str[i] == 'P') { step--;}
        else step++;
        if (step > 0) {
            step = 0;life += mid;;
        }
        if (life+step< 0)return false;
    }
    return true;
}
int main() {
    int t;
    scanf("%d", &t);
    while (t--) {
        scanf("%s", str);
        int len = strlen(str);
        if (str[0] == 'P') { printf("-1\n");continue; }
        int left = 0, right = len, mid;
        int ans;
        while (right >= left) {
            mid = (left + right) / 2;
            if (solve(mid)) {
                ans = mid;right = mid - 1;
            }
            else left = mid + 1;
        }
        printf("%d\n", ans);
    }
    return 0;
}

F - Aggressive cows POJ - 2456  二分

 题目大意：给出n个牛舍坐标（在一直线上），然后有m个牛要入住，现在问相邻牛之间的最小距离最大是几 ，通俗的说就是一条线段上有 n 个点，选取 m 个点，使得相邻点之间的最小距离值最大。

#include <iostream>
#include <map>
#include<cstdio>
#include <string>
#include<algorithm>
#include<vector>
#include<cmath>
#define maxn 100010
int arr[maxn],b[maxn],bj=0,n,c;
using namespace std;
bool solve(int mid) {
    int cnt = 0,add=0;
    for (int i = 0;i < bj;i++) {
        add += b[i];
        if (add >= mid) { cnt++;add = 0; }
    }
    if (cnt >= c - 1)return true;
    return false;
}
int main()
{
    scanf("%d %d", &n, &c);
    for (int i = 0;i < n;i++) {
        scanf("%d", &arr[i]);
    }
    sort(arr, arr + n);
    int right = 0;
    for (int i = 1;i < n;i++) {
        b[bj++] = arr[i] - arr[i - 1];
        right += b[bj - 1];
    }
    int left = 0, mid,res;
    while (right >= left) {
        mid = (right + left) / 2;
        if (solve(mid)) {
            res = mid;left = mid + 1;
        }
        else right = mid - 1;
    }
    printf("%d\n", res);
}

G - 4 Values whose Sum is 0 POJ - 2785 

题目大意：每列选一个看能否组成和为0.

#include <iostream>
#include <map>
#include <string>
#include<algorithm>
#include<vector>
#include<cstdio>
#include<cmath>
using namespace std;
vector<int>a,b,c,d,e,f;
int main()
{
    int n;
    scanf("%d", &n);
    for (int i = 0;i < n;i++) {
        int a1, b1, c1, d1;
        scanf("%d %d %d %d", &a1, &b1, &c1, &d1);
        a.push_back(a1);b.push_back(b1);c.push_back(c1);d.push_back(d1);
    }
    for (int i = 0;i < n;i++) { for (int j = 0;j < n;j++) { e.push_back(a[i] + b[j]); } }
    for (int i = 0;i < n;i++) { for (int j = 0;j < n;j++) { f.push_back(c[i] + d[j]); } }
    sort(f.begin(), f.end());
    int cnt = 0;
    for (int i = 0;i < e.size();i++) {
        int left = 0, right = f.size()-1, mid;
        while (right >= left) {
            mid = (left + right) / 2;
            if (e[i] + f[mid] > 0)right = mid - 1;
            else if (e[i] + f[mid] < 0)left = mid + 1;
            else {
                cnt++;
                for (int j = mid + 1;j < f.size();j++) {
                    if (e[i] + f[j] == 0) { cnt++; }
                    else break;
                }
                for (int j = mid - 1;j >= 0;j--) {
                    if (e[i] + f[j] == 0) { cnt++; }
                    else break;
                }
                break;
            }
        }
    }
    printf("%d\n", cnt);
}