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：
1 #include <cstdio> 2 #include <cstring> 3 #include <algorithm> 4 #define LL long long 5 const long M=110000; 6 LL pos[M],JinWei[M],ans[M],max_pos; 7 long n,t; 8 int main(){ 9 scanf("%d",&t); 10 for (long l=1;l<=t;++l){ 11 scanf("%d",&n); 12 13 memset(pos,0,sizeof(pos)); 14 memset(JinWei,0,sizeof(JinWei)); 15 max_pos=0; 16 long max1=0; 17 while (n--){ 18 long x,a; 19 scanf("%d%d",&a,&x); 20 pos[a]+=x; 21 max1=std::max(max1,a); 22 } 23 24 for (long i=0;i<M;++i) 25 if (pos[i]>1){ 26 pos[i+1]+=(pos[i]>>1); 27 JinWei[i+1]=1; 28 pos[i]=pos[i]%2; 29 } 30 31 max_pos=-1; 32 for (long i=M-1;i>=0;--i) 33 if (pos[i] && !JinWei[i]){ 34 max_pos=i; 35 break; 36 } 37 38 printf("Case #%d: ",l); 39 if (max_pos<0) printf("0 "); 40 else{ 41 memset(ans,0,sizeof(ans)); 42 ans[max_pos]=1; 43 for (long i=max_pos-1;i>=0;--i) 44 if (pos[i]){ 45 ans[max_pos]=0; 46 for (long j=max_pos-1;j>=i;--j) 47 ans[j]=1; 48 max_pos=i; 49 } 50 long output=0; 51 for (long i=M-1;i>=0;--i){ 52 if (ans[i]) output=1; 53 if (output) printf("%d",ans[i]); 54 } 55 printf(" "); 56 } 57 } 58 return 0; 59 }
By 王桢旭