hdu5373

当初做这道题时犯了一个致命的错误。。那就是我一直认为求digit sum就是对9取余。但这道题目不一样，它不要求将digit sum化为一位数，也就是如果当初的数是88，那么digit sum是16，而不是对9取余的7。。
然后还要知道一个数如果能被11整除，它的特点是
奇数位上的和-偶数位上的和能被11整除。
这里是一个链接，列举了一些能被1，2，3……11等整除的数的特征:
整除-百度百科
这样模拟一遍即可。

#include<stdio.h>#include<iostream>#include<string>#include<string.h>#include<algorithm>#include<iomanip>#include<vector>#include<time.h>#include<queue>#include<stack>#include<iterator>#include<math.h>#include<stdlib.h>#include<limits.h>#include<map>#include<set>//#define ONLINE_JUDGE#define eps 1e-8#define INF 0x7fffffff#define FOR(i,a) for((i)=0;i<(a);(i)++)#define MEM(a) (memset((a),0,sizeof(a)))#define sfs(a) scanf("%s",a)#define sf(a) scanf("%d",&a)#define sfI(a) scanf("%I64d",&a)#define pf(a) printf("%d\n",a)#define pfI(a) printf("%I64d\n",a)#define pfs(a) printf("%s\n",a)#define sfd(a,b) scanf("%d%d",&a,&b)#define sft(a,b,num) scanf("%d%d%d",&a,&b,&num)#define for1(i,a,b) for(int i=(a);i<b;i++)#define for2(i,a,b) for(int i=(a);i<=b;i++)#define for3(i,a,b)for(int i=(b);i>=a;i--)#define MEM1(a) memset(a,0,sizeof(a))#define MEM2(a) memset(a,-1,sizeof(a))#define ll __int64const double PI=acos(-1.0);template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}template<class T> inline T Min(T a,T b){return a<b?a:b;}template<class T> inline T Max(T a,T b){return a>b?a:b;}using namespace std;int n,m;#define M 110#define N 100010#define Mod 1000000007#define p(x,y) make_pair(x,y)int getNum(int x){    int num=0;    while(x){        num++;        x/=10;    }    return num;}int main() {#ifndef ONLINE_JUDGE    freopen("in.txt", "r", stdin);//  freopen("out.txt", "w", stdout);#endif    int t;    int kas=1;    while(sfd(n,t)!=EOF){        if(n == -1 && t == -1) break;        int tail = 0;        //保存每次生成的结尾数字        int sum1=0,sum2=0;  //sum1保存奇数位上的和，sum2保存偶数位上的和        int cur=getNum(n);  //当前数字有几位        int tot=cur;        //保存一共有几位        while(n){            if(cur&1) sum1+=(n%10);            else                sum2 += (n%10);            n /= 10;            cur--;        }        tail += sum1+sum2;        int sum = 0;        while(t--){            sum += tail;            int tmp = sum;            tot += getNum(tmp);            cur = tot;            tail=0;            while(tmp){                if(cur&1) sum1+=(tmp%10);                else                    sum2 += (tmp%10);                tail += tmp%10;                tmp /= 10;                cur--;            }        }        int dif = sum1-sum2;        printf("Case #%d: ",kas++);        if(dif%11==0)            printf("Yes\n");        else            printf("No\n");    }    return 0;}