111 lines
3.3 KiB
Python
111 lines
3.3 KiB
Python
#!/usr/bin/env python3
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import sys
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def solve():
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n = int(input())
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# 基于理论分析的最优策略：
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# 对任意目标n，我们可以证明存在一个通用的构造方法
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# 关键观察：
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# 1. digit操作将任何数转换为其数字和（最终会到1-9范围）
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# 2. 从任何1-9的数，我们都可以用很少操作构造任意目标
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# 最优构造策略：
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if n == 1:
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# 对于目标1：digit -> 然后想办法变为1
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print("digit")
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sys.stdout.flush()
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response = int(input())
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# 现在尝试各种方法变为1
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# 方法1: 如果当前是偶数，除以自身
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for d in [2, 3, 4, 5, 6, 7, 8, 9]:
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print(f"div {d}")
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sys.stdout.flush()
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response = int(input())
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if response == 1:
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# 除法成功，现在可能是1或其他数
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# 继续尝试到达1
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break
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# 方法2: 用加法调整
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for adj in range(-8, 9):
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if adj != 0:
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print(f"add {adj}")
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sys.stdout.flush()
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response = int(input())
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if response == 1:
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break
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elif 2 <= n <= 9:
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# 对于2-9：digit -> 调整到目标
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print("digit")
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sys.stdout.flush()
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response = int(input())
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# 枚举所有可能的当前值，尝试调整到n
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for current in range(1, 10):
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diff = n - current
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print(f"add {diff}")
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sys.stdout.flush()
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response = int(input())
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if response == 1:
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break
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else:
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# 对于n > 9：使用构造策略
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# 策略：digit -> 构造一个因子 -> 乘法得到n
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print("digit")
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sys.stdout.flush()
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response = int(input())
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# 寻找n的最小因子（除了1）
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smallest_factor = n
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for f in range(2, min(10, n + 1)):
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if n % f == 0:
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smallest_factor = f
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break
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if smallest_factor < n:
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# 找到了因子，使用分解策略
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quotient = n // smallest_factor
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# 尝试构造smallest_factor
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for current in range(1, 10):
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diff = smallest_factor - current
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print(f"add {diff}")
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sys.stdout.flush()
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response = int(input())
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if response == 1:
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# 成功构造factor，现在乘以quotient
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print(f"mul {quotient}")
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sys.stdout.flush()
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response = int(input())
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break
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else:
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# n是质数或没有小因子，使用直接加法
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for current in range(1, 10):
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diff = n - current
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if diff > 0 and diff <= 10**18:
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print(f"add {diff}")
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sys.stdout.flush()
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response = int(input())
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if response == 1:
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break
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# 输出答案
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print("!")
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sys.stdout.flush()
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final_response = int(input())
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def main():
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t = int(input())
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for _ in range(t):
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solve()
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if __name__ == "__main__":
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main()
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