98 lines
2.8 KiB
Python
98 lines
2.8 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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# 最终策略：基于问题的深入理解
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#
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# 核心思想：
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# 1. 使用digit操作标准化到小范围
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# 2. 利用数学性质构造目标
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# 对于所有情况，先用digit标准化
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print("digit")
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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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return
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# 现在当前值是原数字的数字和
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# 对于1-9的目标，我们有简单策略
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# 对于更大的目标，我们使用构造法
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if n <= 9:
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# 目标是1-9，直接尝试所有可能的调整
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# 由于我们不知道确切的当前值，枚举所有可能性
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attempts = [
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# 假设当前是1-9，计算到n的差值
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n - 1, n - 2, n - 3, n - 4, n - 5,
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n - 6, n - 7, n - 8, n - 9
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]
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for diff in attempts:
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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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# 目标大于9，使用乘法构造
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# 策略：找到n的一个小因子d，使得n = d * q
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# 然后构造d，再乘以q
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best_factor = None
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best_quotient = None
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# 寻找最佳的因子分解
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for d in range(1, 10):
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if n % d == 0:
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q = n // d
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if q <= 10**17: # 确保乘法不会溢出
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best_factor = d
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best_quotient = q
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break
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if best_factor is not None:
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# 使用因子分解策略
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# 先构造best_factor
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for current_guess in range(1, 10):
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diff = best_factor - current_guess
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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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# 成功构造因子，现在乘以quotient
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print(f"mul {best_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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# 没有合适的因子分解，使用直接加法
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# 这种情况下，假设当前是某个小数字，直接加到n
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for current_guess in range(1, 10):
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diff = n - current_guess
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if 0 < 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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# 结束
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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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