Souvenir

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)

Problem Description

Today is the 1st anniversary of BestCoder. Soda, the contest manager, wants to buy a souvenir for each contestant. You can buy the souvenir one by one or set by set in the shop. The price for a souvenir is p yuan
and the price for a set of souvenirs if q yuan.
There‘s m souvenirs
in one set.

There‘s n contestants
in the contest today. Soda wants to know the minimum cost needed to buy a souvenir for each contestant.

Input

There are multiple test cases. The first line of input contains an integer T (1≤T≤105),
indicating the number of test cases. For each test case:

There‘s a line containing 4 integers n,m,p,q (1≤n,m,p,q≤104).

Output

For each test case, output the minimum cost needed.

Sample Input

2
1 2 2 1
1 2 3 4

Sample Output

1
3

Hint

For the first case, Soda can use 1 yuan to buy a set of 2 souvenirs.
For the second case, Soda can use 3 yuan to buy a souvenir.

题目中有两种购买方式，p元1件，或者是q元m件，如果p*m <= q，那么每次都买一件肯定是最小的价格，结果为p*n

如果p*m >q 也就是说如果能买m件，那么花q的钱是优的，所以可以买n/m套，这个价格是便宜的，剩下的n-n/m*m件分两种情况，一种是买n-n/m*m件，一种是买一套，比较一下，选一个小的值。

不能用dp，TLE，，，，

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std ;
#define INF 0x3f3f3f3f
int dp[10010] ;
int main() {
    int t , i , j , ans ,  n , m , p , q ;
    scanf("%d", &t) ;
    while( t-- ) {
        scanf("%d %d %d %d", &n, &m, &p, &q) ;
        if( p*m <= q ) {
            printf("%d\n", n*p) ;
        }
        else {
            i = n/m ;
            ans = i*q ;
            i = n-m*i ;
            if( i*p > q )
                ans += q ;
            else ans += i*p ;
            printf("%d\n", ans) ;
        }
    }
    return 0 ;
}

Hidden String

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)

Problem Description

Today is the 1st anniversary of BestCoder. Soda, the contest manager, gets a string s of
length n.
He wants to find three nonoverlapping substrings s[l1..r1], s[l2..r2], s[l3..r3] that:

1. 1≤l1≤r1<l2≤r2<l3≤r3≤n

2. The concatenation of s[l1..r1], s[l2..r2], s[l3..r3] is
"anniversary".

Input

There are multiple test cases. The first line of input contains an integer T (1≤T≤100),
indicating the number of test cases. For each test case:

There‘s a line containing a string s (1≤|s|≤100) consisting
of lowercase English letters.

Output

For each test case, output "YES" (without the quotes) if Soda can find such thress substrings, otherwise output "NO" (without the quotes).

Sample Input

2
annivddfdersewwefary
nniversarya

Sample Output

YES
NO

直接暴力，枚举对anniversary的两个断点，去和str串匹配

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std ;
int len ;
char str[110] ;
char s[20] = "anniversary" ;
int f(int x,int y) {
    int i , j , k , flag = 0 ;
    for(i = 0 ; i < len ; i++) {
        if( flag == 0 ) {
            for(j = 0 ; j <= x ; j++) {
                if( str[i+j] != s[j] ) break ;
            }
            if( j > x ) {
                flag = 1 ;
                i = i+j-1 ;
            }
        }
        else if( flag == 1 ) {
            for(j = x+1 ; j <= y ; j++) {
                if( str[i+j-x-1] != s[j] ) break ;
            }
            if( j > y ) {
                flag = 2 ;
                i = i+j-x-1-1 ;
            }
        }
        else if( flag == 2 ) {
            for(j = y+1 ; j <= 10 ; j++) {
                if( str[i+j-y-1] != s[j] ) break ;
            }
            if( j > 10 ) {
                return 1 ;
            }
        }
    }
    return 0 ;
}
int main() {
    int t , i , j ;
    scanf("%d", &t) ;
    while( t-- ) {
        scanf("%s", str) ;
        len = strlen(str) ;
        for(i = 0 ; i < 11 ; i++){
            for(j = i+1 ; j < 10 ; j++) {
                if( f(i,j) ) break ;
            }
            if( j < 10 ) break ;
        }
        if( i < 11 )
            printf("YES\n") ;
        else
            printf("NO\n") ;
    }
    return 0 ;
}

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