题目链接：

http://acm.hust.edu.cn/vjudge/problem/113727

Restaurant Ratings

Time Limit: 3000MS

题意

给你一个长度为n，由非负整数组成的和为sum的序列。求长度为也为n，且和比sum小的或者和等于sum并且字典序小于所给序列的所有不同的序列。

题解

数位dp。

dp[i][j]表示序列长度为i时和为j的所有不同序列数。 sumv[i][j]表示长度为i并且和小于等于j的所有不同序列数。

代码

#include<map>

#include<cmath>

#include<queue>

#include<vector>

#include<cstdio>

#include<string>

#include<cstring>

#include<iostream>

#include<algorithm>

using namespace std;

#define X first

#define Y second

#define mkp make_pair

#define lson (o<<1)

#define rson ((o<<1)|1)

#define mid (l+(r-l)/2)

#define sz() size()

#define pb(v) push_back(v)

#define all(o) (o).begin(),(o).end()

#define clr(a,v) memset(a,v,sizeof(a))

#define bug(a) cout<<#a<<" = "<<a<<endl

#define rep(i,a,b) for(int i=a;i<(b);i++)

typedef long long LL;

typedef vector<int> VI;

typedef pair<int,int> PII;

typedef vector<pair<int,int> > VPII;

const int INF=0x3f3f3f3f;

const LL INFL=0x3f3f3f3f3f3f3f3fLL;

const double eps=1e-8;

//start----------------------------------------------------------------------

LL dp[22][33],sumv[22][33];

void pre(){

	clr(dp,0);

	dp[0][0]=1;

	rep(i,1,22){

		rep(j,0,33){

			rep(k,0,j+1){

				dp[i][j]+=dp[i-1][j-k];

			}

			if(j==0) sumv[i][j]=dp[i][j];

			else sumv[i][j]=sumv[i][j-1]+dp[i][j];

		}

	}

}

int arr[22];

int main() {

	pre();

    int n;

    while(scanf("%d",&n)==1&&n){

    	int sum[22];

    	sum[0]=0;

    	rep(i,1,n+1){

    		scanf("%d",&arr[i]);

    		sum[i]=sum[i-1]+arr[i];

		}

		LL ans=0;

    	if(sum[n]-1>=0) ans=sumv[n][sum[n]-1];

    	rep(i,1,n+1){

    		rep(j,0,arr[i]){

    			ans+=dp[n-i][sum[n]-sum[i-1]-j];

			}

		}

		ans++;

		printf("%lld\n",ans);

	}

    return 0;

}

//end-----------------------------------------------------------------------