Good Numbers
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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.
You are required to count the number of good numbers in the range from A to B, inclusive.
Each test case comes with a single line with two numbers A and B (0 <= A <= B <= 1018).
21 101 20
Case #1: 0Case #2: 1
The answer maybe very large, we recommend you to use long long instead of int.思路:不会数位DP,只能找规律。。。在下不才,,找到了一个规律。连续的十个数肯定就会有一个。也就是0-10,有一个,是0,11-20有一个,是19。直接用左右区间极值除以10相减就可以了。但是有一点,举个例子说明一下,127和136,与128和136,如果直接按上述算的话,两组样例结果一样。但是,结果不应该一样,因为在120到130的区间里,有一个符合条件的值是127,在130在140的区间里是136,所一,要判断区间极值在它所在的小区间里的符合条件的值左边还是右边。。(有点绕)。。。代码:#include<iostream>#include<cstdio>#include<cstring>#include<string>#include<map>#include<queue>using namespace std;typedef long long LL;int fun(LL x){ int f[20]; int sum=0,l=0; while(x) { f[l]=x%10; x/=10; sum+=f[l]; l++; } return sum;}int main(){ int T; scanf("%d",&T); int q=1; while(T--) { LL a,b,c=0,d=0,num=0; scanf("%lld%lld",&a,&b); int f1=fun(a/10*10)%10; int s1=0; if(f1!=0) s1=10-f1; if(a%10>s1) num--; int f2=fun(b/10*10)%10; int s2=0; if(f2!=0) s2=10-f2; if(b%10>=s2) num++; c=a/10; d=b/10; printf("Case #%d: %lld\n",q,num+d-c); q++; }}
0 0
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