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hdu4722 Good Numbers

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Good Numbers Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 45 Accepted Submission(s): 14 Problem Description If we sum up every digit of a number and the result can be exactly divided by 10, we say this number is a good number. You are required to count the number of good numbers in the range from A to B, inclusive. Input The first line has a number T (T <= 10000) , indicating the number of test cases. Each test case comes with a single line with two numbers A and B (0 <= A <= B <= 1018). Output For test case X, output "Case #X: " first, then output the number of good numbers in a single line. Sample Input 2 1 10 1 20 Sample Output Case #1: 0 Case #2: 1 Hint The answer maybe very large, we recommend you to use long long instead of int. Source 2013 ACM/ICPC Asia Regional Online —— Warmup2 Recommend zhuyuanchen520 找规律题,水题! #include <iostream> #include <stdio.h> #include <string.h> using namespace std; __int64 ff(__int64 m) { if(m<0) return 0; __int64 temp=m/100,ans; __int64 i; ans=temp*10; for(i=temp*100;i<=m;i++) { __int64 sum=0,t=i; while(t) { sum+=t%10; t/=10; } if(sum%10==0) ans++; } return ans; } int main() { int tcase ,tt=1; __int64 a,b; scanf("%d",&tcase); while(tcase--) { scanf("%I64d%I64d",&a,&b); printf("Case #%d: %I64d\n",tt++,ff(b)-ff(a-1)); } return 0; }

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