HDU 4432 Sum of divisors 2012 Asia Tianjin Regional Contest
2013-05-21 20:54
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解题报告:
题目大意:输入两个数n和m,要求按一下步骤做,首先将n转化成m进制的数,然后求这个数的每一位的平方和。最后转化成m进制再显示出来。
模拟题,要注意的一点是这个m的范围是2到16,所以当m超过10时要记得转化为对应的字母表示。
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题目大意:输入两个数n和m,要求按一下步骤做,首先将n转化成m进制的数,然后求这个数的每一位的平方和。最后转化成m进制再显示出来。
模拟题,要注意的一点是这个m的范围是2到16,所以当m超过10时要记得转化为对应的字母表示。
#include<cstdio> #include<cmath> typedef __int64 ll; ll n,m; char an[17]={'0','1','2','3','4','5','6','7','8','9','A','B','C','D','E','F'}; ll add(ll N,ll M) { ll sum=0; while(N) { ll d=N%M; sum+=d*d; N/=M; } return sum; } void print(ll N,ll M) { char a[10005]; int i=1; while(N) { a[i++]=an[N%M]; N/=M; } for(int j=i-1;j>=1;--j) printf("%c",a[j]); printf("\n"); } int main( ) { while(scanf("%I64d%I64d",&n,&m)!=EOF) { ll sum=0; for(ll i=1;i<=sqrt(n);++i) if(n%i==0) { sum+=add(i,m); if(i*i!=n) sum+=add(n/i,m); } print(sum,m); } return 0; }
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