2014 Multi-University Training Contest 5

官方解题报告http://blog.sina.com.cn/s/blog_a19ad7a10102uyxr.html

Matrix multiplication http://acm.hdu.edu.cn/showproblem.php?pid=4920

#include<cstdio>
#include<cctype>
const int M=810;
int a[M][M],b[M][M],c[M][M];
inline int getint(){
    int res=0;
    char tmp;
    while(!isdigit(tmp=getchar()));
    do{
        res=(res<<3)+(res<<1)+tmp-'0';
    }while(isdigit(tmp=getchar()));
    return res;
}
int main(){
    int n,x,i,j,k;
    while(~scanf("%d",&n)){
        for(i=0;i<n;i++){
            for(j=0;j<n;j++){
                a[i][j]=getint();
                a[i][j]%=3;
            }
        }
        for(i=0;i<n;i++){
            for(j=0;j<n;j++){
                b[i][j]=getint();
                b[i][j]%=3;
                c[i][j]=0;
            }
        }
        for(k=0;k<n;k++){
            for(i=0;i<n;i++){
                if(a[i][k]){
                    for(j=0;j<n;j++){
                        c[i][j]+=a[i][k]*b[k][j];
                    }
                }
            }
        }
        for(i=0;i<n;i++){
            for(j=0;j<n;j++){
                if(j) putchar(' ');
                putchar(c[i][j]%3+'0');
            }
            putchar('\n');
        }
    }
    return 0;
}

 

Inversion http://acm.hdu.edu.cn/showproblem.php?pid=4911

#include<cstdio>
#include<cstring>
#include<cctype>
typedef __int64 LL;
const int M=100010;
int a[M];
class InverseNum {              //逆序对数O(nlogn)
#define _cp(a,b) ((a)<=(b))     //可更改元素类型和比较函数
    typedef int typev;
    typev b[M];
public:
    LL inv(int n,typev a[]) {  //传入序列长度和内容,返回逆序对数
        int l=n>>1,r=n-l,i,j;
        LL ret=(r>1?(inv(l,a)+inv(r,a+l)):0);
        for (i=j=0; i<=l; b[i+j]=a[i],i++)
            for (ret+=j; j<r&&(i==l||!_cp(a[i],a[l+j])); b[i+j]=a[l+j],j++);
        memcpy(a,b,sizeof(typev)*n);
        return ret;
    }
} gx;
int getint(){
    int res=0;
    char tmp;
    while(!isdigit(tmp=getchar()));
    do{
        res=(res<<3)+(res<<1)+tmp-'0';
    }while(isdigit(tmp=getchar()));
    return res;
}
int main(){
    int n,k;
    while(~scanf("%d%d",&n,&k)){
        for(int i=0;i<n;i++){
//            scanf("%d",&a[i]);
            a[i]=getint();
        }
        LL ans=gx.inv(n,a);
        if(ans<k) ans=0;
        else ans-=k;
        printf("%I64d\n",ans);
    }
    return 0;
}

 

 

 

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