Hdu 2089 must be 62-digit dp

Hdu 2089 must be 62-digit dp

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            Using namespace std; int dp [10] [3]; // dp [I] [j] indicates the number of digits, j indicates the status. // dp [I] [0] indicates that an unlucky number does not exist. // dp [I] [1] indicates that the Geely number does not exist, and the highest bit is 2 // dp [I] [2], indicating that there is an unlucky number void init () {memset (dp, 0, sizeof (dp )); dp [0] [0] = 1; // initialize the unlucky number of 0 digits as 1for (int I = 1; I <= 7; I ++) {dp [I] [0] = dp [I-1] [0] * 9-dp [I-1] [1]; // The highest digit does not contain 9 digits of 4, and because when the I-th digit is set to 6, i-1 can not put 2 SO to subtract the I-1 bit does not exist unlucky number to cut the highest bit of 2 dp [I] [1] = dp [I-1] [0]; // The highest bit puts 2 other digits as long as it is not unlucky numbers can dp [I] [2] = dp [I-1] [2] * 10 + dp [I-1] [0] + dp [I-1] [1]; // The number of I-BITs containing unlucky digits is the number of I-1-BITs containing unlucky digits × 10, as long as there is an unlucky number in the back, no matter what the front can be, so the first X 10, and if the highest bit put a 4, then all the following are not unlucky numbers, you can also put the highest bit on a 6 and then put the second to the last bit on 2. It's not unlucky.} int fun (int n) {int len = 0, tem = n, ans, flag, a [10]; while (n) {a [++ len] = n % 10; n/= 10 ;} a [len + 1] = ans = 0; flag = 0; // obtain the unlucky number between 1-(n-1) for (int I = len; i> = 1; I --) {ans + = dp [I-1] [2] * a [I]; // The following I-1 is an unlucky number and then the front can be filled with any if (flag) {ans + = dp [I-1] [0] * a [I];} if (! Flag & a [I]> 4) {ans + = dp [I-1] [0]; // if a [I]> 4, you can put a 4} if (! Flag & a [I + 1] = 6 & a [I]> 2) {// if the last digit is 6, ans + = dp [I] [1];} if (! Flag & a [I]> 6) {ans + = dp [I-1] [1];} if (a [I] = 4 | a [I + 1] = 6 & a [I] = 2) {flag = 1 ;}} return tem-ans;} int main () {int n, m; init (); while (cin> n> m, n | m) {printf ("% d \ n", fun (m + 1)-fun (n);} return 0 ;}