CF 257DIV2 B. Jzzhu and Sequences (取余需注意!!!!!!!)
2014-07-21 13:31
387 查看
题意: f1= x,f2= y, 当i>=2时 fi= f(i-1) + f(i+1)
给你一个x,y,n 求fn%(10e9+7)
题目链接:http://codeforces.com/contest/450/problem/B
算一下就可以发现fn是规律的,fn= f(n-6) 所以我们算出f1- f6即可
但是这题我在比赛的时候被别人hack了 , 原因就是我取余的时候没注意,当ans+MOD 还是负数的时候我就会错
代码:
给你一个x,y,n 求fn%(10e9+7)
题目链接:http://codeforces.com/contest/450/problem/B
算一下就可以发现fn是规律的,fn= f(n-6) 所以我们算出f1- f6即可
但是这题我在比赛的时候被别人hack了 , 原因就是我取余的时候没注意,当ans+MOD 还是负数的时候我就会错
代码:
#include <stdio.h> #include <string.h> #define MOD 1000000007 #define LL long long int main() { LL a[10]; while(scanf("%lld %lld",&a[1], &a[2])!=EOF) { LL n; scanf("%lld",&n); a[3]= a[2]- a[1]; a[4]= a[3]- a[2]; a[5]= a[4]- a[3]; a[6]= a[5]- a[4]; a[7]= a[6]- a[5]; n%= 6; for(int i= 1; i<= 6; i++) { if(i) //之前我是 (a[i]+ MOD)%MOD ,当ans+MOD< 0 会出错 printf("%lld\n",(a[i]%MOD+ MOD)%MOD); else printf("%lld\n",(a[6]%MOD+ MOD)%MOD); } } return 0; }
相关文章推荐
- Jzzhu and Sequences - CF 450B 矩阵快速幂版
- codeforces 257div2 B. Jzzhu and Sequences(细节决定一切)
- CF 450 B Jzzhu and Sequences(矩阵快速幂)
- CodeForces 450B-Jzzhu and Sequences
- codeforces 450-B Jzzhu and Sequences 矩阵快速幂
- (CF#257)B. Jzzhu and Sequences
- Codeforces Jzzhu and Sequences(圆形截面)
- codeforces 450B B. Jzzhu and Sequences(矩阵快速幂)
- Codeforces #257 (Div. 2) B. Jzzhu and Sequences
- Jzzhu and Children - CF 450A 水题
- Jzzhu and Sequences
- Codeforces Round #257 (Div. 2) B. Jzzhu and Sequences
- Codeforces Round #257 (Div. 2) B Jzzhu and Sequences
- codeforces - 450 - B - Jzzhu and Sequences 【简单的矩阵快速幂】
- Codeforces Round #257(Div. 2) B. Jzzhu and Sequences(矩阵快速幂)
- B. Jzzhu and Sequences
- Codeforces Round #257 (Div. 2) B. Jzzhu and Sequences
- Codeforces Round #257 (Div. 2 ) B. Jzzhu and Sequences
- Codeforces Round #257 (Div. 2) B. Jzzhu and Sequences
- CodeForces 450-B. Jzzhu and Sequences