[杂题] Codeforces Gym 101190 NEERC 16 L. List of Primes

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关键是一个函数Solve(int x,int *pre,int sum,int L)
表示已经考虑完了前x个质数得到前缀pre 剩下需要的和为sum 下标从L开始
然后需要预处理下不用前x个质数组成sum 的方案数和总长
然后超出范围的时候不处理
就能过了

#include<cstdio>#include<cstdlib>#include<algorithm>using namespace std;typedef long long ll;const int maxn=7700;int pri[maxn+5],num;int vst[maxn+5];int m=0,n;ll f[1005][10005],g[1005][10005];inline int len(int x){  int ret=0;  while (x) ret++,x=x/10;  return ret+2;}inline void Pre(){  for (int i=2;i<=maxn;i++){    if (!vst[i]) pri[++num]=i;    for (int j=1;j<=num && pri[j]*i<=maxn;j++){      vst[i*pri[j]]=1; if (i%pri[j]==0) break;    }  }  n=num; m=0;  for (int i=1;i<=60;i++) m+=pri[i];  f[n+1][0]=0; g[n+1][0]=1;  for (int i=n;i;i--)    for (int j=0;j<=m;j++){      f[i][j]=f[i+1][j],g[i][j]=g[i+1][j];      if (j>=pri[i]){    f[i][j]+=f[i+1][j-pri[i]]+g[i+1][j-pri[i]]*len(pri[i]);    g[i][j]+=g[i+1][j-pri[i]];      }    }}int first=0; ll start=0;char cur[100005]; int pnt=0;inline void push(int x){  static char tmp[15],len;  len=0;  while (x) tmp[++len]=x%10+'0',x/=10;  for (int i=len;i;i--) cur[++pnt]=tmp[i];  cur[++pnt]=','; cur[++pnt]=' ';}ll L,R;char ans[1000005]; int p;inline void Solve(int x,int sum,ll sta){  if (sta>R) return;  if (sta+f[x+1][sum]+g[x+1][sum]*(2+pnt)<L) return;  if (!sum){    if (!first)      first=1,start=sta+1;    ans[++p]='[';    for (int i=1;i<=pnt-2;i++)      ans[++p]=cur[i];    ans[++p]=']'; ans[++p]=','; ans[++p]=' ';    if (start && start+p-1>=R){      for (ll i=L;i<=R;i++)    putchar(ans[i-start+1]);      exit(0);    }    return;  }  if (x==n) return;  if (f[x+1][sum]==0 || sum<pri[x+1]) return;  int tmp=pnt;  push(pri[x+1]);  Solve(x+1,sum-pri[x+1],sta);  sta+=f[x+2][sum-pri[x+1]]+g[x+2][sum-pri[x+1]]*(2+pnt);  pnt=tmp;  Solve(x+1,sum,sta);}int main(){  freopen("list.in","r",stdin);  freopen("list.out","w",stdout);  scanf("%I64d%I64d",&L,&R);  Pre();  ll t=0;  for (int i=2;;i++){    Solve(0,i,t);    t+=f[1][i]+g[1][i]*2;  }  return 0;}

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