poj3286--How many 0's?(数位dp)
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题目链接:点击打开链接
题目大意:给出m和n(n>=m),求m到n区间内的0的个数和。
数位dp,注意统计前缀为0的,和要求的数的高位存在0,对低位的影响。
#include <cstdio>#include <cstring>#include <algorithm>using namespace std ;#define LL __int64LL dp[12][10] , num[12] ;void init() { int i , j , k ; memset(dp,0,sizeof(dp)) ; dp[1][0] = 1 ; for(i = 2 , num[1] = 1; i < 12 ; i++) num[i] = num[i-1]*10 ; for(i = 2 ; i < 12 ; i++) { dp[i][0] = num[i] ; for(j = 0 ; j < 10 ; j++) { for(k = 0 ; k < 10 ; k++) { dp[i][j] += dp[i-1][k] ; } } } return ;}LL solve(LL x) { int a[12] , cnt = 0 , i , j ; LL sum , s = 0 ; if( x ) sum = 1 ; else sum = 0 ; while( x ) { a[++cnt] = x%10 ; x /= 10 ; } if( !cnt ) a[++cnt] = 0 ; for(i = cnt ; i > 0 ; i--) { if( a[i] == 0 ) s++ ; for(j = 0 ; j < a[i] ; j++) { if( i == cnt && j == 0 ) continue ; sum += dp[i][j] ; sum += s*num[i] ; } if( i == cnt ) continue ; for(j = 1 ; j <= 9 ; j++) sum += dp[i][j] ; } return sum ;}int main() { LL n , m , ans ; init() ; while( scanf("%I64d %I64d", &m, &n) != EOF ) { if( m == -1 && n == -1 ) break ; ans = solve(n) - solve(m) ; do{ if( n%10 == 0 ) ans++ ; n /= 10 ; }while( n ) ; printf("%I64d\n", ans) ; } return 0 ;}
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