5 个海盗分 100 枚金币

import math

def solve(n, G, need):
    """海盗编号 1..m，m 最资深、先提议。need(m) = m 人局通过所需票数（含提议者）。
    偏好：先保命，再多拿金币，无差别时投反对。plan[m] = 分配 dict，None 表示提议者必死。"""
    plan = {1: {1: G}}
    for m in range(2, n + 1):
        k = m - 1
        while plan[k] is None:             # m 死后，往下第一个能活的局面
            k -= 1
        base = plan[k]
        votes = 1 + (m - 1) - k            # 自己 + 夹在中间的必死者（白给的票）
        short = need(m) - votes
        if short <= 0:
            plan[m] = {m: G}; continue
        prices = sorted((base.get(i, 0) + 1, i) for i in range(1, k + 1))
        cost = sum(p for p, _ in prices[:short])
        if cost > G:
            plan[m] = None; continue
        alloc = {i: p for p, i in prices[:short]}
        alloc[m] = G - cost
        alloc['_unique'] = short == len(prices) or prices[short - 1][0] < prices[short][0]
        plan[m] = alloc
    return plan

half = lambda m: math.ceil(m / 2)          # 至少一半（含提议者）
strict = lambda m: m // 2 + 1              # 严格过半

for rule, f in (("至少一半", half), ("严格过半", strict)):
    p = solve(5, 100, f)
    for m in range(2, 6):
        a = p[m]
        if a is None:
            print(rule, m, "提议者必死"); continue
        print(rule, m, [a.get(i, 0) for i in range(m, 0, -1)], "唯一" if a.get('_unique', True) else "不唯一")
# 至少一半 2 [100, 0] 唯一
# 至少一半 3 [99, 0, 1] 唯一
# 至少一半 4 [99, 0, 1, 0] 唯一
# 至少一半 5 [98, 0, 1, 0, 1] 唯一
# 严格过半 2 提议者必死
# 严格过半 3 [100, 0, 0] 唯一
# 严格过半 4 [98, 0, 1, 1] 唯一
# 严格过半 5 [97, 0, 1, 0, 2] 不唯一

p = solve(600, 100, half)
assert all(p[m][m] == 100 - (m - 1) // 2 for m in range(1, 202))
for m in range(2, 201):                    # 方案唯一，且恰好给同奇偶的人各 1 枚
    a = p[m]
    assert a.get('_unique', True)
    assert sorted(i for i in a if i not in ('_unique', m) and a[i] > 0) == \
           [i for i in range(1, m) if i % 2 == m % 2]
print([p[m][m] for m in (5, 10, 50, 199, 200, 201, 202)])   # [98, 96, 76, 1, 1, 0, 0]
print([m for m in range(201, 601) if p[m] is not None])
# [201, 202, 204, 208, 216, 232, 264, 328, 456]