2014.01.15

网址：http://acm.hust.edu.cn/vjudge/contest/view.action?cid=39485#overview

A

题意：看有多少个str字符串里有全部[0,k]个数字

水~

#include<iostream>
#include<cstdio>
#include<string>
using namespace std;

int main(){
    int n,k;
    string str;

    while(cin >> n >> k){
        int ans = 0;
        while(n--){
            cin >> str;
            bool b[10] = {0};

            int len = str.length();
            for (int i = 0; i < len; i++)
                b[str[i]-'0'] = true;
            bool v = true;
            for (int i = 0; i <= k; i++)
                if (!b[i]) v = false;
            if (v) ans++;
        }
        cout <<ans << endl;
    }
}

B

题意：求数列中满足斐波那契的最长长度

没用数组存，滚动的存了下

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

int main(){
     int n;
     int a,b,c;
     int maxm = 0;
     int sum = 0;
     cin >> n;
     if (n > 0) {cin >> a; maxm = ++sum; n--;}
     if (n > 0) {cin >> b; maxm = ++sum; n--;}
     while(n--){
        cin >> c;
        if (c == a + b)
            sum++;
        else{
            sum = 2;
        }
        a = b;
        b = c;
        if (sum > maxm) maxm = sum;
     }
     cout << maxm << endl;
}

C

题意：给了一个a和s，矩阵b[i][j]=s[i]*[j]，求子矩阵个数

第i行是ai*(a1,a2,..,an)，用sum[i]表示数列a的前i项和。

那么任意一个子矩阵和可以表示为(sum[j1]-sum[j0])*(sum[i1]-sum[i0]) ，穷举之。

a==0的情况要特殊考虑

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

int main(){
    int a;
    string s;
    long long v[40000],sum[5000];

    cin >> a >> s;
    int len = s.length();

    memset(sum,0,sizeof(sum));
    memset(v,0,sizeof(v));

    for (int i = 0; i < len; i++)
        sum[i+1] = sum[i] + s[i] - '0';
    for (int i = 1; i <= len; i++)
        for (int j = 0; j < i; j++)
        v[sum[i]-sum[j]]++;
    long long ans = 0L;
    if (a == 0){
            for (int i = 0; i <= sum[len]; i++)
            ans += v[0]*v[i];
    }
    for (int i = 1; i <= sum[len]; i++)
        if (a % i == 0 && a / i <= sum[len])
        ans +=v[a/i]*v[i];

    cout << ans << endl;
}

D

题意：a，b个球分别放在两个盒子里，求最少的步数

用了贪心+卡时，因为a+b<2147483648，算法大约时间是log（a+b）所以卡时碰运气，答题忘写不满足要输出-1了。。。

#include<iostream>
#include<cstdio>
#include<string>
#include<cstring>
typedef long long ll;
using namespace std;
ll a, b;

void swap(ll &a, ll &b){
    if (b < a) {
        ll t = a;
        a = b;
        b = t;
    }
}
int main(){
    cin >> a >> b;

    for (ll i = 0 ; i < 1000000; i++){
        swap(a,b);
        if (a == 0){
            cout << i<< endl;
            return 0;
        }

        b -= a;
        a += a;
    }
    cout << "-1" << endl;
}