codeforces

题目：http://codeforces.com/contest/250/problem/A

简单题目

int main()
{
    int i,j;
    int n;
    int a[110];
    int tnum[110];
    while(scanf("%d",&n) != EOF)
    {
        _clr(tnum,0);
        for(i = 0; i < n; i++)
        scanf("%d",&a[i]);
        int sum = 0;
        int tsum = 0;
        int k = 0;
        for(i = 0; i < n; i++)
        {
            if(a[i] < 0) sum ++;
            tsum ++;
            if(sum == 3)
            {
                //cout<<"i = "<<i<<endl;
                tnum[k++] = tsum - 1;
                tsum = 0;
                sum = 0;
                i --;
            }
        }
        tnum[k ++] = tsum;
        printf("%d\n",k);
        printf("%d",tnum[0]);
        for(i = 1; i < k; i++)
        printf(" %d",tnum[i]);
        cout<<endl;
    }
    return 0;
}

题目：http://codeforces.com/contest/250/problem/B

字符串的处理题目，写的好纠结。先是处理那些两个':'中间那些不够四个的补零，然后如果是两个':'，则保留不做处理，最后补零

const int N = 50;
int main()
{
    char str[N];
    char sbr[N];
    int n,i,j;
    //freopen("data.txt","r",stdin);
    scanf("%d",&n);
    getchar();
    while(n--)
    {
        _clr(str,0);
        _clr(sbr,0);
        int num = 0;
        cin>>str;
        //gets(str);
        //puts(str);
        if(strlen(str) == 2)
        {
            printf("0000:0000:0000:0000:0000:0000:0000:0000\n");
            continue;
        }
        int len = strlen(str);
        for(i = 0; i < len; i++)
        {
            if(str[i] == ':') num ++;
        }
        if(num == 8 && str[0] == ':' && str[1] == ':')
        {
            str[0] = '0';
        }

        if(len == 39)
        {
            printf("%s\n",str);
            continue;
        }
        int sum = 0;
        int k = 0;
        int temp;
        for(i = 0; i < len; i++)
        {
            if(str[i] != ':')
            {
                sum ++;
            }
            else if(str[i - 1] != ':')
            {
                temp = 4 - sum;
                while(temp --) sbr[k ++] = '0';
                for(j = i - sum; j <= i; j++) sbr[k ++] = str[j];
                sum = 0;
            }
            else {sbr[k ++] = ':';sum = 0;}
        }
        if(k == 36 && sbr[k - 1] == ':' && sbr[k - 2] == ':')
        {
            for(i = 0; i < k - 1; i++)
                printf("%c",sbr[i]);
            for(i = 0; i < 4; i++) printf("0");
            printf("\n");
            continue;
        }
        if(sum)
        {
            temp = 4 - sum;
            while(temp --) sbr[k++] = '0';
            for(j = i - sum; j < i; j++) sbr[k ++] = str[j];
        }
    //cout<<"sbr = "<<sbr<<endl;
        //cout<<"sum = "<<k<<endl;
        temp = 39 - k;
        if(sbr[k - 2] == ':' && sbr[k - 1] == ':') temp ++;
        for(i = 0; i < k; i++)
        {
            if(sbr[i] == ':' && sbr[i + 1] == ':')
            {
                printf("%c",sbr[i]);
                for(j = 1; j <= temp; j++)
                {
                    if(j && j % 5 == 0) printf(":");
                    else printf("0");
                }
                i++;
                if(i != (k - 1)) printf(":");
            }
            else printf("%c",sbr[i]);
        }
        cout<<endl;

    }
    return 0;
}