hdu 5248(二分)

题解：二分出最小代价x，然后倒着让每个数字都在x的变化范围内单调递减，如果无法实现就return false，左边界为x + 1，否则右边界是x。

#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
const int N = 100005;
long long n, a[N];

bool judge(long long x) {
    int temp = a[n - 1] + x;
    for (int i = n - 2; i >= 0; i--) {
        if (abs(temp - 1 - a[i]) <= x)
            temp = temp - 1;
        else if (temp - 1 >= a[i])
            temp = a[i] + x;
        else
            return false;
    }
    return true;
}

int main() {
    int t, cas = 1;
    scanf("%d", &t);
    while (t--) {
        scanf("%lld", &n);
        long long l = 0, r = 1000000;
        for (int i = 0; i < n; i++)
            scanf("%lld", &a[i]);


        while (l < r) {
            long long mid = (l + r) / 2;
            if (judge(mid))
                r = mid;
            else
                l = mid + 1;
        }
        printf("Case #%d:\n%lld\n", cas++, l);
    }
    return 0;
}