题目地址：HDU 4588
这题是学长跟我说的1道数位DP。。然后我就按着数位DP去做的，倒是写出来了，但是1直TLE。。后来用类似找规律的方法解出来了。。
首先这题其实就是求每位上总共有多少个1，然后不断从低位开始向高位进位。
方法是比如2进制为1010010的这个数，就能够拆成1000000+10000+103个数，然后从0到111111就是0和1的1个全排序，那末每位上的1的个数总和1定是占1半，然后首位+1，那末10000也能够这样算，然后累加起来就行了。但是还要注意的是，首位的1不只是1000000的时候+1,还有后面的也要加上，所以直接从低位开始枚举，并设置1个变量不断累加就好了。
代码以下：
#include <iostream>
#include <string.h>
#include <math.h>
#include <queue>
#include <algorithm>
#include <stdlib.h>
#include <map>
#include <set>
#include <stdio.h>
using namespace std;
#define LL __int64
#define pi acos(⑴.0)
const int mod=100000000;
const int INF=0x3f3f3f3f;
const double eqs=1e⑻;
int c[40], d[3][40], sum[103];
LL Cal(LL x, int f)
{
int i, cnt=0, j;
LL ans=0;
while(x){
c[cnt++]=(x&1);
x>>=1;
}
for(i=0;i<cnt;i++){
if(c[i]){
d[f][i]+=ans+1;
for(j=0;j<i;j++){
d[f][j]+=(1<<i⑴);
}
ans+=(1<<i);
}
}
}
int main()
{
LL a, b, i, ans;
while(scanf("%I64d%I64d",&a,&b)!=EOF){
memset(sum,0,sizeof(sum));
memset(d,0,sizeof(d));
Cal(b, 0);
if(a)
Cal(a⑴, 1);
ans=0;
for(i=0;i<35;i++){
sum[i]=d[0][i]-d[1][i];
//printf("%I64d
",sum[i]);
}
sum[0]>>=1;
ans=sum[0];
//printf("%I64d
",ans);
for(i=1;i<=100;i++){
sum[i]+=sum[i⑴];
sum[i]>>=1;
ans+=sum[i];
//printf("%I64d
",ans);
}
printf("%I64d
",ans);
}
return 0;
}
#include <string.h>
#include <math.h>
#include <queue>
#include <algorithm>
#include <stdlib.h>
#include <map>
#include <set>
#include <stdio.h>
using namespace std;
#define LL __int64
#define pi acos(⑴.0)
const int mod=100000000;
const int INF=0x3f3f3f3f;
const double eqs=1e⑻;
int c[40], d[3][40], sum[103];
LL Cal(LL x, int f)
{
int i, cnt=0, j;
LL ans=0;
while(x){
c[cnt++]=(x&1);
x>>=1;
}
for(i=0;i<cnt;i++){
if(c[i]){
d[f][i]+=ans+1;
for(j=0;j<i;j++){
d[f][j]+=(1<<i⑴);
}
ans+=(1<<i);
}
}
}
int main()
{
LL a, b, i, ans;
while(scanf("%I64d%I64d",&a,&b)!=EOF){
memset(sum,0,sizeof(sum));
memset(d,0,sizeof(d));
Cal(b, 0);
if(a)
Cal(a⑴, 1);
ans=0;
for(i=0;i<35;i++){
sum[i]=d[0][i]-d[1][i];
//printf("%I64d
",sum[i]);
}
sum[0]>>=1;
ans=sum[0];
//printf("%I64d
",ans);
for(i=1;i<=100;i++){
sum[i]+=sum[i⑴];
sum[i]>>=1;
ans+=sum[i];
//printf("%I64d
",ans);
}
printf("%I64d
",ans);
}
return 0;
}
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波比源码 » HDU 4588 Count The Carries (2013年南京邀请赛)
波比源码 » HDU 4588 Count The Carries (2013年南京邀请赛)
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