题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4734
题意:给定了一个函数,求[0,B]内的数的个数,使得他们换算出来的结果小于F(A)。
数位dp,dp(l,s)表示l位数的值,s为被F(A)减后还剩下的值。如果s<0说明此时的F(x)比F(A)大了,要剪枝掉。直到位数被枚举完全,判断s是否大于等于0。由于这个s最大为9*(∑i=0,9 2^i)≈5120,所以s这一维可以开5120。
1 #include2 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 >> a13 #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 << 121 #define rrt rt << 1 | 122 #define pi 3.1415926535923 #define RT return24 #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 pii;30 typedef pair psi;31 typedef pair pll;32 typedef map msi;33 typedef vector vi;34 typedef vector vl;35 typedef vector vvl;36 typedef vector vb;37 38 const int maxn = 10;39 const int maxm = 5120;40 int digit[maxn];41 int dp[maxn][maxm];42 43 int f(int x) {44 int ret = 0, cnt = 0;45 while(x) {46 ret += (x % 10) * (1< = 0;54 if(s < 0) return 0;55 if(!flag && ~dp[l][s]) return dp[l][s];56 int ret = 0;57 int pos = flag ? digit[l] : 9;58 Rep(i, pos+1) {59 ret += dfs(l-1, s-i*(1<<(l-1)), flag&&(i==pos));60 }61 if(!flag) dp[l][s] = ret;62 return ret;63 }64 65 int solve(int x, int y) {66 int pos = 0;67 while(x) {68 digit[++pos] = x % 10;69 x /=10;70 }71 return dfs(pos, y, true);72 }73 74 int a, b;75 76 signed main() {77 //FRead();78 int T, _ = 1;79 Rint(T);80 Clr(dp, -1);81 W(T) {82 Rint(a); Rint(b);83 a = f(a);84 printf("Case #%d: %d\n", _++, solve(b, a));85 }86 RT 0;87 }