正题

题目链接:http://noip.ybtoj.com.cn/contest/102/problem/1

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题目大意

$n$个$x_i$在$[l_i,r_i]$中随机选择，给出一个$c$，一个序列$\sum k_i=c$
每种方案贡献为$$\prod _{i=1}^n{x}_i^{k_i}$$

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解题思路

设$f_{i,j}$表示到第$i$个小朋友，$c$已经分配了$j$出去时的贡献和
$$f_{i,j}=\sum _{k=0}^j(f_{i,k}+\sum _{z=l_i}^{r_i}{z}^{j-k})$$

然后预处理后面那个东西的前缀和即可，时间复杂度$O(n{c}^2)$

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解题思路

#include<cstdio>
#include<cstring>
#include<algorithm>
#define ll long long
using namespace std;
const ll N=410,XJQ=1e9+7;
ll n,c,l[N],r[N],f[N][N],z[N][N],ans;
signed main()
{freopen("candy.in","r",stdin);freopen("candy.out","w",stdout);scanf("%lld%lld",&n,&c);for(ll i=1;i<=n;i++)scanf("%lld",&l[i]);for(ll i=1;i<=n;i++)scanf("%lld",&r[i]);for(ll i=1;i<=400;i++){z[i][0]=1;for(ll j=1;j<=400;j++)z[i][j]=z[i][j-1]*i%XJQ;}for(ll i=0;i<=400;i++)for(ll j=1;j<=400;j++)(z[j][i]+=z[j-1][i])%=XJQ;f[0][0]=1;for(ll i=1;i<=n;i++)for(ll j=0;j<=c;j++)for(ll k=j;k<=c;k++)(f[i][k]+=f[i-1][j]*(z[r[i]][k-j]-z[l[i]-1][k-j])%XJQ)%=XJQ;printf("%lld",(f[n][c]+XJQ)%XJQ);
}