（赛前练手#2）UVA12620 Fibonacci Sum（找规律。。）

版权声明：转载请声明出处，谢谢支持 https://blog.csdn.net/Dreamstar_DS/article/details/83218618

题目链接：https://www.luogu.org/problemnew/show/UVA12620 OR Vjudge

输入输出样例

输入样例#1： 复制
4
1 3
4 4
5 100
1 99999
输出样例#1： 复制
4
3
5068
4933400

好吧本题可以说是一个提答题（虽然不太恰当），因为模数本题是指定的模数，所以我们直接在本地机器上跑，可以证明对于斐波拉契数列取模的话一定会有循环节，然后就发现循环节是300…，就没有然后了

AC Code:

#include<bits/stdc++.h>
#define rg register
#define maxn 100005
#define il inline
#define ll long long
using namespace std;
il int read(){rg int x = 0  , w = 1;rg char ch = getchar();while (ch < '0' || ch > '9') {if (ch == '-') w = -1;ch = getchar();}while (ch >= '0' && ch <= '9'){x = (x<<3) + (x<<1) + ch - '0';ch = getchar();}return x * w;}
int f[maxn];
int main(){
	rg int t = read();
	ll a , b;
	ll res = 2 , ans = 0;
	f[1] = f[2] = 1;
	for (rg int i = 3 ; i <= 300 ; ++i){
		f[i] = (f[i - 1] + f[i - 2]) % 100;
		res += f[i];
	}
	for (rg int i = 301 ; i <= 3000 ; ++i)
		f[i] = (f[i - 1] + f[i - 2]) % 100;
	while (t--){
		ans = 0;
		scanf("%lld %lld" , &a , &b);
		ll tot = (b - a) / 300;
		ans += tot * res;
		b -= tot * 300;
		ll tota = a / 300;
		a = a - 300 * tota ,  b = b - 300 * tota;
		if (!a) a += 300 , b += 300;
		for (rg int i = a ; i <= b ; ++i) ans += f[i];
		printf("%lld\n" , ans);
	}
	return 0;	
}```