20160815训练记录

T1

题解：枚举两边长。。会多出来一小截。。XJB统计一下

#include<cmath>
#include<cstdio>
#include<iostream>
using namespace std;
int s2k(int k){
    return k*(k-1)/2;
}
int ans=0;
int main(){
    freopen("rectangle.in","r",stdin);
    freopen("rectangle.out","w",stdout);
    int n,m,k;
    scanf("%d %d %d",&n,&m,&k);
    if(n>m)swap(n,m);
    int a=min((int)sqrt((double)k),n);
    
    for(int i=2;i<=a;i++){
        int b=min(m,k/i);
        if(k<b*(i+1)){
            if(b<m)
                ans=max(ans,s2k(i)*s2k(b)+s2k(k-i*b)*b);
            else
                ans=max(ans,s2k(i)*s2k(b)+s2k(k-i*b)*i);
        }            
    }
    cout<<ans<<endl;
    return 0;
}

T2

题解：然后两个前缀和相减就能算了

用矩阵快速幂计算

#include<cstdio>
#include<string.h>
#include<iostream>
#define mod 10000
using namespace std;
int f[1000005];
struct mat{
    int a[2][2];
    int*operator [](int _a){
        return a[_a];
    }
};
mat c,sple;
mat mul(mat a,mat b){
    mat t;
    memset(t.a,0,sizeof(t.a)); 
    for(int i=0;i<2;i++)for(int j=0;j<2;j++)for(int k=0;k<2;k++)t[i][j]=(t[i][j]+a[i][k]*b[k][j])%10000;
    return t;
}
int intp(int a,int n){
    if(!n)return 1;
    int _p=intp(a,n/2);
    return _p*_p*((n&1)?a:1);
}
mat matp(mat a,long long n){
    if(!n)return c;
    mat t=matp(a,n/2);
    return mul(mul(t,t),(n&1ll?a:c));
}
int getfib(long long n){
    mat a; 
    a[0][0]=1;a[0][1]=1;
    a[1][0]=1;a[1][1]=0;
    for(int i=0;i<2;i++)for(int j=0;j<2;j++)c[i][j]=(i==j);
    mat ans=matp(a,n);    
    //printf("(%d %d\n%d %d)\n",ans[0][0],ans[0][1],ans[1][0],ans[1][1]);
    return ans[0][1];
}
int main(){
    freopen("fibonacci.in","r",stdin);
    freopen("fibonacci.out","w",stdout);
    int T;cin>>T; 
    //f[1]=1;f[2]=1;for(int i=3;i<=1000;i++)f[i]=(f[i-1]+f[i-2])%mod;
    //for(int i=1;i<=1000;i++)printf("%d\n",f[i]);
    while(T--){
        //f(y+2)-f(x+1)-2
        int x,y;scanf("%d %d",&x,&y);
        
        printf("%d\n",((getfib((long long)y+2)-getfib((long long)x+1))%mod+mod)%mod);
        //printf("force ans=%d\n",(f[y+2]-f[x+1]+mod)%mod);
    }
    return 0;
}