C :

#include "stdio.h"
#include "math.h"
int i,n,k,T;
double a,b,c,d;
int f[21];
void print_1(int s)
{
	int a[4],d;
	if(n==0)
	{
		printf("0\n");
		return ;
	}
	d=0;
	while(s)
		a[d++]=s%10,s/=10;
	for(i=d-1;i>=0&&d-i<=k;i--)
		printf("%d",a[i]);
	printf("\n");
}
void print_2()
{
	c=log10(a)+n*log10(b);
	c-=(int)c;
	d=pow(10,c);
	while(d<1000)
		d*=10;
	print_1((int)d);
}
int main()
{
	f[0]=0;
	f[1]=f[2]=1;
	for(i=3;i<=20;i++)
		f[i]=f[i-1]+f[i-2];
	a=sqrt(5);
	b=(1+a)/2;
	a/=5;
	scanf("%d",&T);
	while(T--)
	{
		scanf("%d%d",&n,&k);
		if(n<=20)
			print_1(f[n]);
		else
			print_2();
	}
	return 0;
}

C++ :

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <vector>
#include <climits>
#include <stack>
#include <queue>
#include <map>
#include <algorithm>
#include <string>
#include <cstring>
#define MID(x,y) ((x+y)>>1)
const double f=(sqrt(5.0)+1.0)/2.0;
using namespace std;
typedef long long LL;

int fab[21];
int main(){
    //freopen("test.in", "r", stdin);
    //freopen("test.out", "w", stdout);
    int t;
    scanf("%d", &t);
    while(t --){
        int n, k;
        fab[0] = 0;
        fab[1] = 1;
        for (int i = 2; i <= 20; i ++)
            fab[i] = fab[i-1] + fab[i-2];
        cin >> n >> k;
        double log10_fabn;
        if (n <= 20){
            if (fab[n] <= pow(10, k-1)){
                cout << fab[n] << endl;
                continue;
            }
            else{
                log10_fabn = log(fab[n]);
            }
        }
        else{
            log10_fabn = -0.5*log(5.0)/log(10.0)+((double)n)*log(f)/log(10.0);//忽略最后一项无穷小
        }
        log10_fabn = log10_fabn - (int)log10_fabn;
        double num = pow(10, log10_fabn);
        while(num < pow(10, k-1))
            num *= 10;
        cout << (int)num <<endl;
    }
	return 0;
}