Windows 10

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解法：首先可以保证每次相同长度的操作不会超过一次，因为如果选择两次那么选择当前长度+1的操作会更优，然后每次都贪心搜索。先预处理出每种长度的操作所下降的值，然后每次都贪心找到尽量接近的数来操作，然后每次都判一下选择比当前数大于等于的操作是否更优，记录操作的步数以及更换的次数。注意音量小于0的情况是直接变为0的。
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#include <iostream>#include <stdio.h>#include <stdlib.h>using namespace std;const int INF = 1e9;long long a[50] = {0}, x, y, k, ans, t, count, point;long long max(long long x, long long y) {    if (x > y) return x;    return y;}long long min(long long x, long long y) {    if (x < y) return x;    return y;}int main() {    int tt;    scanf("%d", &tt);    t = 1;    for (int i = 1; i <= 32; i++) {        t = t*2;        a[i] = t-1;    }    while (tt--) {        scanf("%lld %lld", &x, &y);        if (x <= y) {            printf("%lld\n", y-x);            continue;        }        t = x-y;        k = 0;        ans = INF;        count = 0;        while (1) {            point = 0;            for (int i = 31; i >= 0; i--) if (a[i] < t) {                point = i;                break;            }            if (k+count+max(0, a[point+1]-t-k)+point+1 < ans)                ans = k+count+max(0, min(y,a[point+1]-t-k))+point+1;            if (point == 0) break;                      count = count+point;            k = k+1;            t = t-a[point];        }        printf("%lld\n", ans);    }}