紫书第七章例题（暴力求解法）

1.除法（Division, UVa 725）

#include <cstdio>
#include <cstring>
bool isOk(int a, int b){
    char buff[20];
    int visited[15];
    memset(visited, 0, sizeof(visited));
    sprintf(buff, "%05d%05d", a, b);
    for (int i =0; i < 10; i++){
        int temp = buff[i] - '0';
        if (0 == visited[temp]){
            visited[temp]++;
        } else {
            return false;
        }
    }
    return true;
}
int main(void){
    int N, cnt = 0;
    bool flag = false;
    for (;scanf("%d", &N) && N;){
        if (cnt > 0){
            printf("\n");
        }
        cnt++;
        flag = false;
        for (int i = 1234; i< 100000; i++){
            if (0 == i%N){
                int ans = i/N;
                if (isOk(i, ans)){
                    printf("%05d / %05d = %d\n", i, ans, N);
                    flag = true;
                } 
            }
        }
        if (!flag){
             printf("There are no solutions for %d.\n",N);
        }        
    }
    return 0;
}

PE了，真是神奇，，C++处理字符串还是不够熟练。这种简单题目不能一次过就说明还是太水啊！！

2.最大乘积（Maximum Product, UVa 11059）

#include <cstdio>
#include <cstring>
#define ll long long
int a[20], n, cnt;
int main(void){
    for (cnt = 1;~scanf("%d", &n);cnt++){
        for (int i = 1; i <= n; i++){
            scanf("%d", &a[i]);
        }
        ll S = 1, maxn = 0;
        for (int i = 1; i <= n; i++){
            S = 1;//忘记
            for (int j = i; j <= n; j++){
                S *= a[j];
                if (S > maxn){
                    maxn = S;
                }
            }
        }
        if (maxn < 0){
            maxn = 0;
        }
        printf("Case #%d: The maximum product is %lld.\n\n", cnt, maxn);
        getchar();
    }
    return 0;
}

WA，因为逻辑没理清，忘这忘那！！

 3.素数环（Prime Ring Problem, UVa 524）

#include "cstdio"
#include "cstring"
#include "cmath"
int n,vis[20], A[20];
bool isPrime[50];

void dfs(int cur){
    if (cur==n && isPrime[A[0]+A[n-1]]) {
        for (int i=0; i < n-1; i++)    printf("%d ", A[i]);
        printf("%d\n", A[n-1]);
    } else {
        for (int i = 2; i <= n; i++) {
            if (!vis[i] && isPrime[i+A[cur-1]]) {
                A[cur] = i;
                vis[i] = 1;
                dfs(cur+1);
                vis[i] = 0;
            }
        }
    }
}
void check(){
    for (int i = 2; i < 50; i++){
        isPrime[i] = true;
        for (int k = 2; k < sqrt(i)+1; k++){
            if (i%k == 0){
                isPrime[i] = false;
                break;
            }
        }
    }
}
int main(int argc, char const *argv[])
{
    int t = 0;
    check();
    for (;~scanf("%d", &n);){
        memset(vis, 0, sizeof(vis));
        A[0] = 1;
        if (t > 0){printf("\n");}
        printf("Case %d:\n", ++t);
        dfs(1);
    }
    return 0;
}

这一题PE两次，空行输出真是个大坑。

 4.困难的串（Krypton Factor, UVa 129）

#include <cstdio>
#include <cstring>
#include <string>

int n,L, cnt;
char v[81];

bool judge(int cur)
{
    for(int i=1;i<=(cur+1)/2;i++)//判断长度为2*i的后缀是否有相同的
    {
        bool equal=1;
        for(int j=0;j<i;j++)
            if(v[cur-i-j]!=v[cur-j])  
            {
                equal=0;
                break;
            }
        if(equal)
            return false;
    }
    return true;
}

int dfs(int cur)
{
    for(int i=0;i<L;i++)
    {
        v[cur]='A'+i;
        if(judge(cur))
        {
            cnt++;
            if(cnt==n)
            {
                 //attention format
                 for (int i=0; i<=cur; i++){
                     printf("%c", v[i]);
                     if ((i+1)%4 == 0 && i!=cur) {
                         printf(" ");
                     }
                     if (i!=cur && (i+1)%64==0){
                         printf("\n");
                     }
                 }
                 printf("\n%d", cur+1);
                return 1;
            }
            if(dfs(cur+1))
                return 1;
        }
    }
    return 0;
}

int main()
{
    for(;scanf("%d%d", &n, &L);) {
        cnt=0;
        dfs(0);    
    }
    return 0;
}

超时！

#include "cstdio"
int n, L, cnt;
int S[100];
int dfs(int cur){ 
    if(cnt++ == n){
        for(int i = 0; i < cur; i++) {
            printf("%c", 'A'+S[i]);
            if ((i+1)!=cur && (i+1)%4 ==0 && (i+1)!=64){printf(" ");}
            if ((i+1)!=cur && (i+1)%64 == 0){printf("\n");}
        }
        printf("\n%d\n", cur);
    return 0;
    } 
    for(int i = 0; i < L; i++){
        S[cur] = i;
        int ok = 1;
        for(int j = 1; j*2 <= cur+1; j++){
            int equal = 1;
            for(int k = 0; k < j; k++)
                if(S[cur-k] != S[cur-k-j]) { equal = 0; break; }
            if(equal) { ok = 0; break; }
        } 
        if(ok) if(!dfs(cur+1)) return 0;
    }
    return 1;
}
int main(int argc, char const *argv[])
{
    for (;scanf("%d%d", &n, &L) && n;){
        cnt = 0;
        dfs(0);
    }
    return 0;
}

PE三次，更可怕的是思路也被完全碾压，回溯！！