任务 · 算法 中等 Python ·
Ways to make change
Count the ways to make amount from coins of the given distinct positive denominations, each usable any number of times. Order doesn't matter (1+2 and 2+1 are the same way). Return the count modulo 1,000,000,007. There is exactly one way to make 0. Up to 60 denominations and amounts up to 30,000.
函数签名
def coin_ways(coins: list[int], amount: int) -> int:示例
coin_ways([1, 2, 5], 5) # -> 4
coin_ways([2], 3) # -> 0
coin_ways([10], 10) # -> 1另有 7 个隐藏测试和 1 个限时海量数据测试；其输入数据不公开。 任务以英文提示词提供给模型。
第 1 名Mistral Devstral 2 2512已解决 · 通过 11/11 个测试生成耗时 906 毫秒 · 执行耗时 345 毫秒 · $0.0003 ✓ coin_ways([1, 2, 5], 5)预期 4 2 毫秒 ✓ coin_ways([2], 3)预期 0 <1 毫秒 ✓ coin_ways([10], 10)预期 1 <1 毫秒 ✓ <1 毫秒✓ <1 毫秒✓ <1 毫秒✓ <1 毫秒✓ <1 毫秒✓ <1 毫秒✓ 15 毫秒⏱ 322 毫秒代码
def coin_ways(coins: list[int], amount: int) -> int: MOD = 10**9 + 7 dp = [0] * (amount + 1) dp[0] = 1 for coin in coins: for i in range(coin, amount + 1): dp[i] = (dp[i] + dp[i - coin]) % MOD return dp[amount]第 1 名Qwen3 Coder Next已解决 · 通过 11/11 个测试生成耗时 2.8 秒 · 执行耗时 352 毫秒 · $0.0002 ✓ coin_ways([1, 2, 5], 5)预期 4 <1 毫秒 ✓ coin_ways([2], 3)预期 0 <1 毫秒 ✓ coin_ways([10], 10)预期 1 2 毫秒 ✓ 1 毫秒✓ <1 毫秒✓ 2 毫秒✓ 3 毫秒✓ <1 毫秒✓ <1 毫秒✓ 14 毫秒⏱ 327 毫秒代码
def coin_ways(coins: list[int], amount: int) -> int: MOD = 1000000007 # dp[i] = number of ways to make amount i dp = [0] * (amount + 1) dp[0] = 1 # One way to make 0: use no coins # For each coin, update dp array for all amounts >= coin for coin in coins: for i in range(coin, amount + 1): dp[i] = (dp[i] + dp[i - coin]) % MOD return dp[amount]第 1 名SpaceXAI Grok Build 0.1已解决 · 通过 11/11 个测试生成耗时 5.4 秒 · 执行耗时 354 毫秒 · $0.0022 ✓ coin_ways([1, 2, 5], 5)预期 4 <1 毫秒 ✓ coin_ways([2], 3)预期 0 <1 毫秒 ✓ coin_ways([10], 10)预期 1 <1 毫秒 ✓ <1 毫秒✓ <1 毫秒✓ 1 毫秒✓ <1 毫秒✓ <1 毫秒✓ <1 毫秒✓ 14 毫秒⏱ 334 毫秒代码
def coin_ways(coins: list[int], amount: int) -> int: MOD = 1000000007 dp = [0] * (amount + 1) dp[0] = 1 for coin in coins: for x in range(coin, amount + 1): dp[x] = (dp[x] + dp[x - coin]) % MOD return dp[amount]第 1 名Anthropic Claude Haiku 5.5已解决 · 通过 11/11 个测试生成耗时 1.7 秒 · 执行耗时 355 毫秒 · $0.0002 ✓ coin_ways([1, 2, 5], 5)预期 4 <1 毫秒 ✓ coin_ways([2], 3)预期 0 2 毫秒 ✓ coin_ways([10], 10)预期 1 <1 毫秒 ✓ <1 毫秒✓ <1 毫秒✓ 1 毫秒✓ 1 毫秒✓ 1 毫秒✓ 1 毫秒✓ 15 毫秒⏱ 330 毫秒代码
def coin_ways(coins: list[int], amount: int) -> int: MOD = 1_000_000_007 if amount < 0: return 0 dp = [0] * (amount + 1) dp[0] = 1 for c in coins: if c <= 0 or c > amount: continue for a in range(c, amount + 1): dp[a] = (dp[a] + dp[a - c]) % MOD return dp[amount] % MOD
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