首先我們很容易想到乙個dp。 讓f
[i][
j]' role="presentation" style="position: relative">f[i
][j]
f[i]
[j]表示
i' role="presentation" style="position: relative">i
i次操作後有
j' role="presentation" style="position: relative">j
j個紅球的拿球的方案數。
易得轉移 若j
>
0' role="presentation" style="position: relative">j
>0j
>0，
f[i]
[j]−
>f[
i+1]
[j−1
]' role="presentation" style="position: relative">f[i
][j]
−>f[
i+1]
[j−1
]f[i
][j]
−>f[
i+1]
[j−1
] f[
i][j
]−>f[
i+1]
[j]' role="presentation" style="position: relative">f[i
][j]
−>f[
i+1]
[j]f
[i][
j]−>f[
i+1]
[j]
若j<
n' role="presentation" style="position: relative">j
f[i]
[j]−
>f[
i+1]
[j+1
]' role="presentation" style="position: relative">f[i
][j]
−>f[
i+1]
[j+1
]f[i
][j]
−>f[
i+1]
[j+1
] f[
i][j
]−>f[
i+1]
[j]' role="presentation" style="position: relative">f[i
][j]
−>f[
i+1]
[j]f
[i][
j]−>f[
i+1]
[j]
但是這樣子我們會算重複，因為詢問的是顏色序列的方案數，不是拿球的方案數。
觀察發現答案為共
n' role="presentation" style="position: relative">n
n個球的答案減去共n−
1' role="presentation" style="position: relative">n−1
n−1個球的答案，於是好像沒了？
怎麼證？我也不知道qaq
#include
#include
#include
#include
using
namespace
std;
const
int n=3005,mod=1000000007;
int n,m,f[n][n*2];
int calc(int n)
for(int i=0;ifor(int j=0;j<=n;j++)
if(j1][j+1]=(f[i+1][j+1]+f[i][j])%mod;
f[i+1][j]=(f[i+1][j]+f[i][j])%mod;}}
}int ans=0;
for(int i=0;i<=n;i++)
return ans;
}int main()