UVA11361 Investigating Div-Sum Property
Investigating Div-Sum Property
题目大意:给定一个k,求a <= x <= b的,且x%k为0,x十进制下%k为0,有多少个x
题解:
数位dp套路题
ddp[i][j][a]表示i位数,%k为j,各位数加和%k为a的数有多少
初始化出i=1的状态,枚举往后面一位加的数b dp[i+1][(j * 10 + b)%k][(j+b)%k] += dp[i][j][a]
直接求a与b之间的不好求,转为求1....b减去1....a-1
从高位到低位枚举,比如13245:
枚举0****,10***,11***,12***,130**,131**,1320*,1321*,1322*,1323*,分别用对应状态加入答案
然后枚举13240,13241,13242,13243,13244,13245,加入答案
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <cstdlib> 5 #include <algorithm> 6 #include <queue> 7 #include <vector> 8 #define min(a, b) ((a) < (b) ? (a) : (b)) 9 #define max(a, b) ((a) > (b) ? (a) : (b)) 10 #define abs(a) ((a) < 0 ? (-1 * (a)) : (a)) 11 inline void swap(long long &a, long long &b) 12 { 13 long long tmp = a;a = b;b = tmp; 14 } 15 inline void read(long long &x) 16 { 17 x = 0;char ch = getchar(), c = ch; 18 while(ch < '0' || ch > '9') c = ch, ch = getchar(); 19 while(ch <= '9' && ch >= '0') x = x * 10 + ch - '0', ch = getchar(); 20 if(c == '-') x = -x; 21 } 22 23 const long long INF = 0x3f3f3f3f; 24 const long long MAXK = 100 + 10; 25 26 long long t,a,b,k,dp[40][MAXK][MAXK],powp[40],n,tmp; 27 //dp[i][j][a]表示i位数,这个数%k=j,各位数之和%k=a 28 29 30 //求1...m有多少满足条件的数 31 long long solve(long long m) 32 { 33 long long now = 0, sum = 0, ans = 0, ma = 0; 34 while(powp[ma] <= m) ++ ma; 35 for(register long long i = ma;i >= 2;-- i) 36 { 37 while(now + powp[i - 1] <= m) 38 { 39 ans += dp[i - 1][(k - (now % k))%k][(k - (sum % k))%k]; 40 now += powp[i - 1]; 41 ++ sum; 42 } 43 } 44 for(;now <= m;++ now, ++ sum) 45 if(now % k == 0 && sum % k == 0) ++ ans; 46 return ans; 47 } 48 49 int main() 50 { 51 powp[0] = 1; 52 for(register long long i = 1;i <= 35;++ i)powp[i] = powp[i - 1] * 10; 53 read(t); 54 for(;t;--t) 55 { 56 read(a), read(b), read(k); 57 if(k > 100) 58 { 59 printf("0\n"); 60 continue; 61 } 62 memset(dp, 0, sizeof(dp)); 63 n = 0; 64 while(powp[n] <= b) ++ n; 65 for(register long long i = 0;i <= 9;++ i) ++ dp[1][i % k][i % k]; 66 for(register long long i = 1;i < n;++ i) 67 for(register long long j = 0;j < k;++ j) 68 for(register long long a = 0;a < k;++ a) 69 for(register long long b = 0;b <= 9;++ b) 70 dp[i + 1][(j * 10 + b) % k][(a + b) % k] += dp[i][j][a]; 71 printf("%lld\n", solve(b) - solve(a - 1)); 72 } 73 return 0; 74 }