• [URAL1057]Amount of Degrees(数位dp)


    题目链接:http://acm.timus.ru/problem.aspx?space=1&num=1057

    题意:求[L,R]区间内的数转换成b进制后可以变成k个b进制的幂和。

    其实就是求区间内的数里有几个数,使得这些数转换成b进制有k个1存在。

    转换成对应进制,然后dfs。每一位插0或者1就行了。(这题其实前导零不用处理)

     1 #include <bits/stdc++.h>
     2 using namespace std;
     3 #define fr first
     4 #define sc second
     5 #define cl clear
     6 #define BUG puts("here!!!")
     7 #define W(a) while(a--)
     8 #define pb(a) push_back(a)
     9 #define Rint(a) scanf("%d", &a)
    10 #define Rll(a) scanf("%I64d", &a)
    11 #define Rs(a) scanf("%s", a)
    12 #define Cin(a) cin >> a
    13 #define FRead() freopen("in", "r", stdin)
    14 #define FWrite() freopen("out", "w", stdout)
    15 #define Rep(i, len) for(int i = 0; i < (len); i++)
    16 #define For(i, a, len) for(int i = (a); i < (len); i++)
    17 #define Cls(a) memset((a), 0, sizeof(a))
    18 #define Clr(a, x) memset((a), (x), sizeof(a))
    19 #define Full(a) memset((a), 0x7f7f7f, sizeof(a))
    20 #define lrt rt << 1
    21 #define rrt rt << 1 | 1
    22 #define pi 3.14159265359
    23 #define RT return
    24 #define lowbit(x) x & (-x)
    25 #define onecnt(x) __builtin_popcount(x)
    26 typedef long long LL;
    27 typedef long double LD;
    28 typedef unsigned long long ULL;
    29 typedef pair<int, int> pii;
    30 typedef pair<string, int> psi;
    31 typedef pair<LL, LL> pll;
    32 typedef map<string, int> msi;
    33 typedef vector<int> vi;
    34 typedef vector<LL> vl;
    35 typedef vector<vl> vvl;
    36 typedef vector<bool> vb;
    37 
    38 const int maxn = 33;
    39 int dp[maxn][maxn][2];
    40 int digit[maxn];
    41 int x, y, k, b;
    42 
    43 int mul(int x, int n) {
    44   int ret = 1;
    45   while(n) {
    46     if(n & 1) ret *= x;
    47     n >>= 1; x *= x;
    48   }
    49   return ret;
    50 }
    51 
    52 int dfs(int l, int cnt, bool fz, bool flag) {
    53   if(l == 0) return cnt == k;
    54   if(!flag && ~dp[l][cnt][fz]) return dp[l][cnt][fz];
    55   int ret = 0;
    56   int pos = flag ? digit[l] : b-1;
    57   Rep(i, min(pos+1, 2)) {
    58     if(i==0 && fz) ret += dfs(l-1, cnt, true, flag&&(i==pos));
    59     else ret += dfs(l-1, cnt+i, fz&&(i==0), flag&&(i==pos));
    60   }
    61   if(!flag) dp[l][cnt][fz] = ret;
    62   return ret;
    63 }
    64 
    65 int f(int t) {
    66   int pos = 0;
    67   while(t) {
    68     digit[++pos] = t % b;
    69     t /= b;
    70   }
    71   return dfs(pos, 0, true, true);
    72 }
    73 
    74 signed main() {
    75   //FRead();
    76   Clr(dp, -1);
    77   while(~scanf("%d%d%d%d",&x,&y,&k,&b)) {
    78     printf("%d
    ", f(y) - f(x-1));
    79   }
    80   RT 0;
    81 }
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  • 原文地址:https://www.cnblogs.com/kirai/p/5897938.html
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