D. Divide

题目链接：戳这里

思路：用一个数组记录下每个二进制位上的值，满2进位，从高位到低位扫一遍，找到不能被均分的最高位（当前位为1而且不是进位获得的），该位置代表的值减去比它低位的值就是答案了

AC Code：

#include <cstdio>
#include <cstring>
#include <algorithm>
#define LL long long
const long M=110000;
LL pos[M],JinWei[M],ans[M],max_pos;
long n,t;
int main(){
    scanf("%d",&t);
    for (long l=1;l<=t;++l){
        scanf("%d",&n);

        memset(pos,0,sizeof(pos));
        memset(JinWei,0,sizeof(JinWei));
        max_pos=0;
        long max1=0;
        while (n--){
            long x,a;
            scanf("%d%d",&a,&x);
            pos[a]+=x;
            max1=std::max(max1,a);
        }

        for (long i=0;i<M;++i)
        if (pos[i]>1){
            pos[i+1]+=(pos[i]>>1);
            JinWei[i+1]=1;
            pos[i]=pos[i]%2;
        }

        max_pos=-1;
        for (long i=M-1;i>=0;--i)
            if (pos[i] && !JinWei[i]){
                max_pos=i;
                break;
            }

        printf("Case #%d: ",l); 
        if (max_pos<0) printf("0
");
        else{
            memset(ans,0,sizeof(ans));
            ans[max_pos]=1;
            for (long i=max_pos-1;i>=0;--i)
                if (pos[i]){
                    ans[max_pos]=0;
                    for (long j=max_pos-1;j>=i;--j)
                        ans[j]=1;
                    max_pos=i;
                }
        long output=0;
        for (long i=M-1;i>=0;--i){
            if (ans[i]) output=1;
            if (output) printf("%d",ans[i]);
        }
        printf("
");
        }
    }
    return 0;
}

By 王桢旭