Codeforces Round #370(div 2)

A B C :=w=

D:两个人得分互不影响很关键

一种是f[i][j]表示前i轮，分差为j的方案数

明显有f[i][j]=f[i-1][j-2k]+2*f[i-1][j-2k+1]+...+(2k+1)f[i-1][j]+...

枚举是K*T*T，转移是K 超时

不过发现转移的时候可以理解为j是j-1的算式的整个区间移动，于是可以维护f[i-1]的前缀和来达到O(1)的转移

另一种f[i][j]表示前i轮，某个人分数为j的方案数(两人互不影响，无先后手之分，所以两人等价）

式子写出来就少了第一种的前面的系数，还是要维护前缀和，更加清楚直接了

最后统计时候枚举第一个人的分数，累乘结果

E：？？？

时间： 2024-10-09 09:04:17

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