【NYOJ222】整数中的

#include<stdio.h>int a[256]={ 0,1,1,2,1,2,2,3,1,2,2,3,2,3,3,4,1,2,2,3,2,3,3,4,2,3,3,4,3,4,4,5, 1,2,2,3,2,3,3,4,2,3,3,4,3,4,4,5,2,3,3,4,3,4,4,5,3,4,4,5,4,5,5,6, 1,2,2,3,2,3,3,4,2,3,3,4,3,4,4,5,2,3,3,4,3,4,4,5,3,4,4,5,4,5,5,6, 2,3,3,4,3,4,4,5,3,4,4,5,4,5,5,6,3,4,4,5,4,5,5,6,4,5,5,6,5,6,6,7, 1,2,2,3,2,3,3,4,2,3,3,4,3,4,4,5,2,3,3,4,3,4,4,5,3,4,4,5,4,5,5,6, 2,3,3,4,3,4,4,5,3,4,4,5,4,5,5,6,3,4,4,5,4,5,5,6,4,5,5,6,5,6,6,7, 2,3,3,4,3,4,4,5,3,4,4,5,4,5,5,6,3,4,4,5,4,5,5,6,4,5,5,6,5,6,6,7, 3,4,4,5,4,5,5,6,4,5,5,6,5,6,6,7,4,5,5,6,5,6,6,7,5,6,6,7,6,7,7,8};int main(){ /*freopen("Input.txt","r",stdin); freopen("2.txt","w",stdout);*/ //freopen("2.txt","r",stdin); int n=0,m=0,k=0; //scanf("%d",&n); while(scanf("%d%d",&m,&k)!=EOF) { //scanf("%d",&m); /*m=256-1-n; m=n-256-1; m=m^(m>>1); m=m^(m>>2); m=m^(m>>4); m=m^(m>>8); m=m^(m>>16); printf("%d/n",m&1); m=(m&0x55555555)+((m>> 1)&0x55555555); m=(m&0x33333333)+((m>> 2)&0x33333333); m=(m&0x0f0f0f0f)+((m>> 4)&0x0f0f0f0f); m=(m&0x00ff00ff)+((m>> 8)&0x00ff00ff); m=(m&0x0000ffff)+((m>>16)&0x0000ffff); printf("%d/n",m);*/ for(;m<=k;m++) { unsigned char *ptr=(unsigned char *)&m; n+=(a[ptr[0]]+a[ptr[1]]+a[ptr[2]]+a[ptr[3]]); } printf("%d/n",n); } return 0;} 

 

下边的是数学推理出的方法：

#include<cstdio>unsigned f[]={0,1,4,12,32,80,192,448,1024,2304,5120,11264,24576,53248,114688,245760,524288,1114112,2359296,4980736,10485760,22020096,46137344,96468992,201326592,419430400,872415232,1811939328};int getpos(int n){ int cnt=0; while(n) { cnt++; n>>=1; } return cnt;}int count1(int num){ int i=0; while(num) { num&=(num-1); i++; } return i;}int getsub(int s){ s|=s>>1; s|=s>>2; s|=s>>4; s|=s>>8; s|=s>>16; return s=~s;}int g(int m){ if(!(m&(m-1))) { int n=getpos(m); return f[n-1]; } return g((m&(-m)))+g((m&(m-1)))+count1((getsub(m&(-m))&(m&(m-1))))*(m&(-m));}int main(){ int a,b; scanf("%d%d",&a,&b); printf("%d",g(b+1)-g(a));}