30 lines
679 B
C++
30 lines
679 B
C++
#include<cstdio>
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#include<iostream>
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#include<cstring>
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using namespace std;
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unsigned long long t,a[101],x,y;
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unsigned long long find(unsigned long long now)//求区间1和now之间有多少个1
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{
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unsigned long long ans=0,j;//初始化
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while (now)//还可以拆分
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{
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j=1;
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while (a[j+2]<=now) j++;//寻找可以减的
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ans+=a[j];//加上一的个数
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now-=a[j+1];//减（拆分）
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}
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return ans;
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}
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int main()
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{
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scanf("%llu",&t);//读入
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a[1]=1; a[2]=1;//初始化
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for (int i=3;i<=100;i++)
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a[i]=a[i-1]+a[i-2];//斐波那契
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for (unsigned long long i=1;i<=t;i++)
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{
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scanf("%llu%llu",&x,&y);//读入
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printf("%llu\n",find(y)-find(x-1));//利用前缀和求出答案，然后输出
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}
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return 0;
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} |