poj 1953(斐波那契数列)

#include<iostream>
#include<cstdio>
using namespace std;
int main(){
    int t,i,kase;
    long long n,data[50];
    data[1] = 2;
    data[2] = 3;
    for(i=3;i<=45;i++){
        data[i] = data[i-1]+data[i-2];
    }
    scanf("%d",&t);
    kase = 1;
    while(kase<=t){
        scanf("%lld",&n);
        printf("Scenario #%d:\n",kase++);
        printf("%lld\n\n",data[n]);
    }
    return 0;
}

 

posted @ 2021-08-09 15:57  智人心  阅读(52)  评论(0)    收藏  举报