Chapter2二分与前缀和

Chapter 2 二分与前缀和

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• 二分

• 套路

如果更新方式写的是R = mid， 则不用做任何处理，如果更新方式写的是L = mid，则需要在计算mid是加上1。

1.数的范围 789

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
//整数二分
using namespace std;

int st[100005];
int n, q;
int u;

int main()
{
    scanf("%d%d", &n, &q);
    for(int i = 0; i < n; i++)
        scanf("%d", &st[i]);
    while(q--)
    {
        int L = 0; int R = n - 1;
        scanf("%d", &u);
        while(L < R)
        {
            int md = L + R >> 1;
            if(st[md] >= u) R = md;//注意边界问题
            else L = md + 1;
        }
        if(st[R] == u)
        {
            cout << R << " ";
            R = n - 1;
            while(L < R)
            {
                int md = L + R + 1 >> 1;//注意边界问题
                if(st[md] <= u) L = md;
                else R = md - 1;
            }
            cout << L << endl;
        }
        else    cout << "-1 -1" << endl;
    }
    return 0;
}

2.数的三次方根 790

#include <iostream>
//实数二分
using namespace std;

int main()
{
    double n;
    cin >> n;
    double l = -10000, r = 10000;
    while(r - l > 1e-8)
    {
        double mid = (l + r) / 2;
        if(mid * mid * mid >= n)   r = mid;
        else l = mid;
    }
    printf("%f
", l);
    return 0;
}

3.机器人跳跃问题 730

//AC code
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

const int maxn = 1e5 + 5;
int h[maxn], temp, n;

int main()
{
    cin >> n;
    for(int i = 0; i < n; i++)
        scanf("%d", h + i);
    int l = 0, r = 1e5;
    while(l < r)
    {
        int mid =(l + r) >> 1;//注意边界问题
        temp = mid;
        for(int i = 0; i < n && mid >= 0; i++)
        {
            mid = mid * 2 - h[i];
            if(mid >= 1e5)  break;   //没有这句ac不了，防止中间过程爆掉
        }
        if(mid >= 0)
            r = temp;
        else
            l = temp + 1;
    }
    cout << r << endl;
    return 0;
}

4.四平方和 1221

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

const int maxn = 5e6;//注意
int n, m;

struct Sum{
    int s, c, d;
    
    bool operator< (const Sum &t)const
    {
        if(s != t.s)    return s < t.s;
        if(c != t.c)    return c < t.c;
        return d < t.d;
    }
}p[maxn];

int main()
{
    cin >> n;
    for(int c = 0; c * c <= n; c++ )
        for(int d = c; d * d + c * c <= n; d++ )
            p[m++] = {c * c + d * d, c, d};
    
    sort(p, p + m);
    
    for(int a = 0; a * a <= n; a++ )
        for(int b = a; b * b + a * a <= n; b++ )
        {
            int t = n - a * a - b * b;
            int l = 0, r = m - 1;
            while(l < r)
            {
                int mid = (l + r) >> 1;
                if(p[mid].s >= t)  r = mid;
                else l = mid + 1;
            }
            if(p[l].s == t)
            {
                printf("%d %d %d %d
", a, b, p[l].c, p[l].d);
                return 0;
            }
        }
}

5.分巧克力 1227

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

const int maxn = 100010;
int n, k;
int h[maxn], w[maxn];

bool check(int u)
{
    int sum = 0;
    for(int i = 0; i < n; i++)
        sum += (h[i] / u) * (w[i] / u);
    if(sum >= k)    return true;
    return false;
}

int main()
{
    cin >> n >> k;
    for(int i = 0; i < n; i++)
        scanf("%d%d", h + i, w + i);
    int L = 1, R = 1e5;
    while(L < R)
    {
        int mid = L + R + 1 >> 1;
        if(check(mid))  L = mid;
        else R = mid - 1;
    }
    cout << R << endl;
    return 0;
}

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• 前缀和

1.前缀和 795

//一维数组前缀和
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

const int maxn = 100010;

int a[maxn], s[maxn];
int n, m;

int main()
{
    cin >> n >> m;
    for(int i = 1; i <= n; i++ )
    {
        scanf("%d", a + i);
        s[i] = s[i - 1] + a[i];//存储前i个数的和
    }
    while(m--)
    {
        int x, y;
        scanf("%d%d", &x, &y);
        printf("%d
", s[y] - s[x - 1]);
    }
    return 0;
}

2.子矩阵的和 796

// 二维数组前缀和
#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>

using namespace std;

const int maxn = 1010;
int a[maxn][maxn], s[maxn][maxn];
int n, m, q;

int main()
{
    scanf("%d%d%d", &n, &m, &q);
    for(int i = 1; i <= n; i++ )
        for(int j = 1; j <= m; j++ )
        {
            scanf("%d", &a[i][j]);
            s[i][j] = s[i - 1][j] + s[i][j - 1] - s[i - 1][j - 1] + a[i][j];
        }
    while(q--)
    {
        int x1, y1, x2, y2;
        scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
        printf("%d
", s[x2][y2] - s[x1 - 1][y2] - s[x2][y1 - 1] + s[x1 - 1][y1 - 1]);
    }
    return 0;
}

3.激光炸弹 99

4.K倍区间 1230

//优化时间复杂度

//O(n)
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

typedef long long LL;

const int maxn = 100010;
int s[maxn], cnt[maxn];
int n, k, temp;
LL res;

int main()
{
    cnt[0] = 1;
    cin >> n >> k;
    for(int i = 1; i <= n; i++ )
    {
        scanf("%d", &temp);
        s[i] = (s[i - 1] + temp) % k;
        res += cnt[s[i]];
        cnt[s[i]]++;
    }
    
    printf("%lld
", res);
    return 0;
}

K倍区间学习到的经验：

1.首先因为知道考的是前缀和，我就用O(n²)的复杂度的二重for循环暴力枚举，果不其然超时，然后就没有了进一步的简化思路。

2.好的思路是从简单暴力的方法中优化出来的，就如本题的AC code，把复杂度降到了O(n)，成功AC。

3.找到右端点后遍历前面所有前缀和，符合K倍区间性质的其实就是mod k后与右端点前缀和mod k后余数相等的点。（★）