[GCJKickstart2018RoundB] A. No Nine

题目链接：https://code.google.com/codejam/contest/10284486/dashboard#s=p0

求$[L,R]$内不被9整除且每一位都不是9的数的个数。

经典数位dp，根据$(x+y)\%m=(x\%m+y\%m)\%m$，直接维护前$l$位对9取模的值，向后dfs到最后一位发现模数不是0就计数。

复制

#include <bits/stdc++.h>
using namespace std;

using LL = long long;
const int maxn = 22;
int digit[maxn];
LL dp[maxn][maxn];
LL L, R;

LL dfs(int l, int mod, bool flag) {
  if(!flag && ~dp[l][mod]) return dp[l][mod];
  if(l == 0) {
    if(mod % 9 == 0) return dp[l][mod] = 0;
    return dp[l][mod] = 1;
  }
  int pos = flag ? digit[l] : 9;
  LL ret = 0;
  for(int i = 0; i <= pos; i++) {
    if(i == 9) continue;
    ret += dfs(l-1, (mod*10+i)%9, flag&&(i==pos));
  }
  if(!flag) dp[l][mod] = ret;
  return ret;
}

LL f(LL x) {
  int tot = 0;
  while(x) {
    digit[++tot] = x % 10;
    x /= 10;
  }
  return dfs(tot, 0, true);
}

signed main() {
  freopen("in", "r", stdin);
  freopen("out", "w", stdout);
  int T, _ = 1;
  scanf("%d", &T);
  memset(dp, -1, sizeof(dp));
  while(T--) {
    scanf("%lld%lld",&L,&R);
    printf("Case #%d: %lld\n", _++, f(R)-f(L-1));
  }
  return 0;
}