题目链接：http://lightoj.com/volume_showproblem.php?problem=1140
给出区间[m,n]，求区间内的所有数共有多少个0。
设dp[i][j]表示处理到第i位时，它前面共有j个0（除了前导零）。
注意：前导0
1 #include <algorithm> 2 #include <iostream> 3 #include <cstring> 4 #include <cstdlib> 5 #include <cstdio> 6 #include <vector> 7 #include <ctime> 8 #include <queue> 9 #include <list> 10 #include <set> 11 #include <map> 12 using namespace std; 13 #define INF 0x3f3f3f3f 14 typedef long long LL; 15 16 LL dp[12][12]; 17 int bit[12]; 18 LL dfs(int len, int zer, int sum, int flag) 19 { 20 if(len < 1){ 21 if(zer == 0) 22 return 1; 23 return sum; 24 } 25 if(dp[len][sum] != -1 && flag && zer)//注意zer 26 return dp[len][sum]; 27 int te = flag ? 9 : bit[len]; 28 LL res = 0; 29 for(int i = 0; i <= te; i++) 30 { 31 int Sum = sum, Zer = zer; 32 if(zer == 0 && i != 0) 33 Zer = 1; 34 if(i == 0 && zer == 1) 35 Sum++; 36 res += dfs(len-1, Zer, Sum, flag || i < te); 37 } 38 if(flag && zer)//注意zer,一定要写 39 dp[len][sum] = res; 40 return res; 41 } 42 LL solve(LL n) 43 { 44 int len = 0; 45 memset(bit, 0, sizeof(bit)); 46 while(n) 47 { 48 bit[++len] = n % 10; 49 n /= 10; 50 } 51 return dfs(len, 0, 0, 0); 52 } 53 int main() 54 { 55 int t; 56 LL l, r; 57 memset(dp, -1, sizeof(dp)); 58 scanf("%d", &t); 59 for(int ca = 1; ca <= t; ca++) 60 { 61 scanf("%lld %lld", &l, &r); 62 printf("Case %d: %lld ", ca, solve(r)-solve(l-1)); 63 } 64 return 0; 65 }