UVALive - 6872 Restaurant Ratings 数位dp

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

include

include

include

include

include

include

include

include

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 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-----------------------------------------------------------------------