c语言二分法 & 1ll用法

二分法思想很简单，举个例子。

0--100 random一个数字num

让你猜这个数字是多少？

第一步肯定先猜50~，这就是二分的思想。好了，到此为止，你已经学会了二分法，接下来让我们做到题练习一下。

思路：

1，遍历时间：从min（1） ~ max(rank[i] * cars  * cars)

2,   时间肯定超了，那就用二分法，一半一半猜~

show show code

bool check(long long time,int *ranks, int ranksSize,int cars)
{
    long long  m = 0;
    for(int i = 0;i<ranksSize;i++)
    {
        m +=    sqrt(time/ranks[i]);
    }
    return m >=  cars;
}
typedef long long ll;
long long repairCars(int* ranks, int ranksSize, int cars){
    long long res = 0;
    long long left = 1, right =  1ll * ranks[0] * cars * cars;
    while(left < right)
    {
        long long mid = (left + right)>>1;
        if(check(mid,ranks,ranksSize,cars))
        {
            right = mid;
        }else{
            left = mid+1;
        }
    } 
    return left;

}

bool get(long long budget,int num,int* comp,int *stock,int stockSize,int* cost,int costSize)
{
    for(int i = 0;i<stockSize;i++)
    {
        long long temp = 1ll*(1ll*num*comp[i] - stock[i]);
        if(temp > 0)
        {
            budget -= (long long)temp*cost[i];
        }
    }
    return budget >= 0;
}

int money(int n ,long long budget, int* comp,int *stock,int stockSize,int* cost,int costSize)
{
    int right = (budget/cost[0] + stock[0])/comp[0];
    int left = 0;
    while(left < right)
    {
        int mid = left + (right-left)/2;
        if(get(budget,mid,comp,stock,stockSize,cost,costSize))
        {
            left = mid + 1;
        }else
        {
            right = mid - 1;
        }
    }
    if(get(budget,left,comp,stock,stockSize,cost,costSize))
    {
        return left;
    }
    return left-1;
}


int maxNumberOfAlloys(int n, int k, int budget, int** composition, int compositionSize, int* compositionColSize, int* stock, int stockSize, int* cost, int costSize){
    int res = 0;


    for(int i = 0;i<compositionSize;i++)
    {
        res = fmax(res,money(n,(long long)budget,composition[i],stock,stockSize,cost,costSize));
    }
    return res;
}