http://acm.nyist.net/JudgeOnline/problem.php?pid=301&&递推求值
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矩阵运算。。。这一题让我明白了一些事,在做题的时候一定要考虑数的取值范围。。。否者会多吃WA的这一题我就是因为这wa了好几次。。。
#include<iostream>#include<string.h>#include<cstdio>#include<algorithm>#define M 1000007typedef long long L;typedef struct{ L s[3][3];}Node;Node a,b;Node ceil(Node p,Node q){ Node c; memset(c.s,0,sizeof(c.s)); for(int i=0;i<3;++i) for(int j=0;j<3;++j) for(int t=0;t<3;++t) c.s[i][j]=((c.s[i][j]+p.s[i][t]*q.s[t][j])%M+M)%M; return c;}Node doit(int k){ Node p=b,q=a; while(k){ if(k&1) p=ceil(p,q); q=ceil(q,q); k=k>>1;}return p;}int main(){ int T; scanf("%d",&T);while(T--){ int f1,f2,a1,b1,c1,n; scanf("%d%d%d%d%d%d",&f1,&f2,&a1,&b1,&c1,&n);if(n==1) printf("%d\n",f1); else if(n==2) printf("%d\n",f2); else if(n>=3) { a.s[0][0]=0;a.s[0][1]=a1;a.s[0][2]=0;a.s[1][0]=1;a.s[1][1]=b1;a.s[1][2]=0;a.s[2][0]=0;a.s[2][1]=1;a.s[2][2]=1; b.s[0][0]=1;b.s[0][1]=0;b.s[0][2]=0;b.s[1][0]=0; b.s[1][1]=1;b.s[1][2]=0;b.s[2][0]=0;b.s[2][1]=0;b.s[2][2]=1; n-=2; Node c=doit(n); L k=((f1*c.s[0][1]+f2*c.s[1][1]+c1*c.s[2][1])%M+M)%M; printf("%lld\n",k);}}return 0;}
一般一个题这样算下来不超过1亿的话,一秒左右都能过;
嗯,大致数量级就是亿级别
十亿就很可能会挂
不过,这都只是一般情况
;
编译器不同,机器配置不同,操作系统不同;
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