Bipartite summary! !

One-half of the batch foolish!

Today even half of Nima wrote several hair wa

Easy error occurred:

1, the right border is not enough. (Beginning 1e18 been wa, 2e18 off)

2, logic errors.

/*
几种二分方法整理
元素可以重复 
*/
 
//lower_bound(num, num+size, x)-num:大于等于x的第一个数的下标 
//upper_bound(num, num+size, x)-num:大于x的第一个数的下标 
 
//1.求等于x的最小的index，不存在返回-1
int binary (int *num, int start, int end, int x) {
    int l = start, r = end, ans=-1;
    while(l <= r) {
        int mid = (l+r) >> 1;
        if(num[mid] == x) {
            ans = mid;
            r = mid - 1;
        }
        else if(num[mid] > x)
            r = mid - 1;
        else
            l = mid + 1;
    }
    return ans;
}
 
//2.求等于x的最大的index，不存在返回-1
int binary (int *num, int start, int end, int x) {
    int l = start, r = end, ans=-1;
    while(l <= r) {
        int mid = (l+r) >> 1;
        if(num[mid] == x) {
            ans = mid;
            l = mid + 1;
        }
        else if(num[mid] > x)
            r = mid - 1;
        else
            l = mid + 1;
    }
    return ans;
}
 
//3.求小于x的最大的index
int binary (int *num, int start, int end, int x) {
    int l = start, r = end;
    while(l <= r) {
        int mid = (l+r) >> 1;
        if(num[mid] >= x)
            r = mid - 1;
        else
            l = mid + 1;
    }
    return r;
}
 
//4.求大于x的最小的index
int binary (int *num, int start, int end, int x) {
    int l = start, r = end;
    while(l <= r) {
        int mid = (l+r) >> 1;
        if(num[mid] <= x)
            l = mid + 1;
        else
            r = mid - 1;
    }
    return l;
}
 
//5.求大于等于x的最小的index
int binary (int *num, int start, int end, int x) {
    int l = start, r = end;
    while(l <= r) {
        int mid = (l+r) >> 1;
        if(num[mid] >= x)
            r = mid - 1;
        else
            l = mid + 1;
    }
    return l;
}
 
//6.求小于等于x的最大的index
int binary (int *num, int start, int end, int x) {
    int l = start, r = end;
    while(l <= r) {
        int mid = (l+r) >> 1;
        if(num[mid] <= x)
            l = mid + 1;
        else
            r = mid - 1;
    }
    return r;
}

Today foolish batch ++