CQUPT Training Dec4rd - Dec10th

题目链接

F - Linearization of the kernel functions in SVM

题意

书写表达式
f(x,y,z) = ap + bq + cr + du + ev + fw + gx + hy + iz + j
有0，1，-1等特殊情况自行处理

思路

满是坑坑的题目，特殊情况好多...
-1 0 0 0 0 0 0 0 0 0
1 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 1
0 0 0 0 0 0 0 0 0 -1
等等..

AC代码

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

int main(){
    int t;
    char s[100] = {"pqruvwxyz\0"};
    scanf("%d",&t);
    bool first = true;
    while( t-- ){
        int l;
        first = true;
        for( int i = 0 ; i < 9 ; i++ ){
            scanf("%d",&l);
            if( l > 0 && !first){
                if( l != 1 )
                    printf("+%d%c",l,s[i]);
                else
                    printf("+%c",s[i]);
            }else if( l < 0 || (first && l != 0) ){
                first = false;
                if( l != 1 && l != -1 )
                    printf("%d%c",l,s[i]);
                else if( l == 1 )
                    printf("%c",s[i]);
                else
                    printf("-%c",s[i]);
            }
        }
        scanf("%d",&l);
        if( l > 0 && !first )
            printf("+");
        if( l != 0 || first )
            printf("%d",l);
        printf("\n");
    }
    return 0;
} 

P - Friendship of Frog

题意

找字符串中两个相同字符距离的最小值，没有相同的输出-1

思路

记录出现字符过的最新位置，每次更新ans和字符位置即可

AC代码

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cstdlib> 
#include <cmath> 
using namespace std;
int fcount[50],now[50],ans;
char s[1007];

void cc(){
    memset(fcount,-1,sizeof(fcount));
    memset(now,-1,sizeof(now));
    ans = -1;
}

int main(){
    int t,cas = 0,len;
    scanf("%d",&t);
    while(t--){
        scanf("%s",s);
        len = strlen(s);
        cc();
        for( int i = 0 ; i < len ; i++ ){
            int f = s[i] - 'a';
            if( now[f] == -1 ){
                now[f] = i;
            }else{
                if( fcount[f] == -1 || fcount[f] > i - now[f]  ){
                    fcount[f] = i - now[f];
                }
                now[f] = i;
                if( fcount[f] != -1 && (ans == -1 || fcount[f] < ans ) ){
                    ans = fcount[f];
                }
            }
        }
        printf("Case #%d: %d\n",++cas,ans);
    }
    return 0;
}