int a = 12345678;
//格式为sring输出
Label1.Text = string.Format("asdfadsf{0}adsfasdf",a);
Label2.Text = "asdfadsf"+a.ToString()+"adsfasdf";
Label1.Text = string.Format("asdfadsf{0:C}adsfasdf",a);//asdfadsf￥1,234.00adsfasdf
Label2.Text = "asdfadsf"+a.ToString("C")+"adsfasdf";//asdfadsf￥1,234.00adsfasdf
double b = 1234.12543;
int a = 12345678;
//格式为特殊的string样式输出
Label1.Text = string.Format("asdfadsf{0:C}adsfasdf",b);//asdfadsf￥1,234.13adsfasdf
Label2.Text = "asdfadsf"+b.ToString("C")+"adsfasdf";//asdfadsf￥1,234.13adsfasdf
Label1.Text = string.Format("{0:C3}",b);//￥1,234.125
Label2.Text = b.ToString("C3");//￥1,234.125
Label1.Text = string.Format("{0:d}",a);//十进制--12345678
Label2.Text = b.ToString("d");//十进制--相同的类型，转换报错
Label1.Text = string.Format("{0:e}",a);//指数--1.234568e+007
Label2.Text = b.ToString("e");//指数--1.234125e+003
Label1.Text = string.Format("{0:f}",a);//定点数--12345678.00
Label2.Text = b.ToString("f");//定点数--1234.13
Label1.Text = string.Format("{0:n}",a);//数值--12,345,678.00
Label2.Text = b.ToString("n");//数值--1,234.13
Label1.Text = string.Format("{0:x}",a);//十六进制--bc614e
Label2.Text = b.ToString("x");//16--带有小数不能转换，出错
Label1.Text = string.Format("{0:g}",a);//通用为最紧凑--12345678
Label2.Text = b.ToString("g");//通用为最紧凑--1234.12543
Label1.Text = string.Format("{0:r}",a);//转来转去不损失精度--整数不允许用，报错
Label2.Text = b.ToString("r");//转来转去不损失精度--1234.12543
double b = 4321.12543;
int a = 1234;
自定义模式输出：
//"0"描述：占位符，如果可能，填充位
Label1.Text = string.Format("{0:000000}",a);// 001234
Label2.Text = string.Format("{0:000000}",b);// 004321
//"#"描述：占位符，如果可能，填充位
Label1.Text = string.Format("{0:####### }",a);// 1234
Label2.Text = string.Format("{0:####### }",b);// 4321
Label1.Text = string.Format("{0:#0#### }",a);// 01234
Label2.Text = string.Format("{0:0#0000}",b);// 004321
//"."描述：小数点
Label1.Text = string.Format("{0:000.000}",a);//1234.000
Label2.Text = string.Format("{0:000.000}",b);//4321.125
double b = 87654321.12543;
int a = 12345678;
//","描述：数字分组，也用于增倍器
Label1.Text = string.Format("{0:0,00}",a);// 12,345,678
Label2.Text = string.Format("{0:0,00}",b);// 87,654,32
Label1.Text = string.Format("{0:0,}",a);// 12346
Label2.Text = string.Format("{0:0,}",b);// 87654
Label1.Text = string.Format("{0:0,,}",a);// 12
Label2.Text = string.Format("{0:0,,}",b);// 88
Label1.Text = string.Format("{0:0,,,}",a);// 0
Label2.Text = string.Format("{0:0,,,}",b);// 0
//"%"描述：格式为百分数
Label1.Text = string.Format("{ 0:0% }",a);// 1234567800%
Label2.Text = string.Format("{ 0:#% }",b);// 8765432113%
Label1.Text = string.Format("{ 0:0.00% }",a);// 1234567800.00%
Label2.Text = string.Format("{ 0:#.00% }",b);// 8765432112.54%
//"abc"描述：显示单引号内的文本
Label1.Text = string.Format("{0:'文本'0}",a);// 文本12345678
Label2.Text = string.Format("{0:文本0}",b);// 文本87654321
//"\"描述：后跟1要打印字的字符，也用于转移符\n等
Label1.Text = string.Format("\"你好！\"");// "你好！"
Label2.Text = string.Format("[url=file://\\c\\books\\new\\we.asp]\\c\\books\\new\\we.asp");//\c\books\new\we.asp
//"@"描述：后跟要打印字的字符,
Label1.Text = string.Format(@"""你好！"""); // "你好！"要打印"则需要输入两对才可以
Label2.Text = string.Format(@"\c\books\new\we.asp");//\c\books\new\we.asp 

题目链接：http://202.120.106.94/onlinejudge/problemshow.php?pro_id=143

这道题嘛，怎么说呢，好吧中等题

要求算出下山的前k短路的路长度

由于一定是下山所以可以用邻接表记录路径，然后用一个优先队列记录已有的到n的路长度

题目链接：http://acm.pku.edu.cn/JudgeOnline/problem?id=3659

这题不算难题了，基本算是中等题

题目大意是给出一颗树，在一些点建一个信号塔，信号塔覆盖范围是其所在点和邻近点，问最少几个信号塔可以覆盖全区域

这次比赛成绩比预期差

开始Ultramanhu调整IDE

Q Boy从头开始看题

我的任务是倒数看题，最后看的题目是J，I，H，G

我看完J觉得J可做（哈密顿回路），但是需要很长时间。就首先放着继续看题

树状数组模块

ACM个人模板

POJ 2155 题目测试通过

/**
 * 树状数组模块
 * 下标从0开始
 */
typedef long DG_Ran;
typedef long DG_Num;
const DG_Num DG_MAXN = 1005;

//2^n
DG_Num LowBit(DG_Num n)
{
    return n & (-n);
}
//获取父节点索引
DG_Num DGFather(DG_Num n)
{
    return n + LowBit(n + 1);
}
//获取小的兄弟节点索引
DG_Num DGBrother(DG_Num n)
{
    return n - LowBit(n + 1);
}
//查找增加树状数组前pos项和
//参数(树状数组[in],索引[in],初始赋0即查找前n项和[out])
//复杂度:log(n)
void DGFind(DG_Ran *g,DG_Num pos,DG_Ran &sum)
{
    sum += *(g + pos);
    if(pos >= LowBit(pos + 1))
        DGFind(g, pos - LowBit(pos + 1), sum);
}
//查找对应线性数组元素
//参数(树状数组[in],索引[in]).
//返回值:对应线性数组元素log(n)
//复杂度:log(n)
DG_Ran DGFindEle(DG_Ran *g,DG_Num pos)
{
    DG_Ran a = 0 , b = 0;
    DGFind(g, pos, a);
    if(pos)
    {
        DGFind(g,pos - 1,b);
        return a - b;
    }
    else
        return a;
}
//树状数组,增加节点
//参数:树状数组[out],原数组大小[in],新增线性数组值[in]
//复杂度:log(n)
DG_Ran DGAdd(DG_Ran *g,DG_Num n,DG_Ran val)
{
    *(g + n) = val;
    DG_Num a = n;
    DG_Num b = 1;
    while((a & (~b)) != a)
    {
        *(g + n) += *(g + a - 1);
        a &= (~b);
        b <<= 1;
    }
    return n + 1;
}
//构建树状数组
//参数:线性数组[in],数组大小[in],树状数组[out]
//复杂度:nlog(n)
DG_Ran DGCreate(DG_Ran *g,DG_Num n,DG_Ran *tg)
{
    DG_Num i;
    *tg = *g;
    for(i = 1 ; i < n ; i ++)
        DGAdd(tg,i,*(g + i));
    return n;
}
//修改指定位置值
//参数:线性数组[in],数组位置[in],数组大小[in],新值[in]
//复杂度:log(n)
DG_Ran DGEdit(DG_Ran *g,DG_Num pos,DG_Num n,DG_Ran val)
{
    DG_Num f = DGFather(pos);
    DG_Ran o = *( g + pos );
    *( g + pos ) = val;
    if(f < n)
    {
        DG_Ran fv = val - o + *( g + f );
        DGEdit(g, f, n, fv);
    }
    return n;
}

//树状数组的翻转(树状数组的应用)
//一维  复杂度log(n)
//小于等于指定位置的元素的翻转<=pos
void DGDown1(DG_Ran g[],DG_Num pos,DG_Ran av)
{
    while(pos >= 0)
        g[pos] += av , pos = DGBrother(pos);
}
//获取位置pos的元素翻转次数
DG_Ran DGCUp1(DG_Ran g[],DG_Num pos , DG_Num n)
{
    DG_Ran t = 0;
    while(pos < n)
        t += g[pos] , pos = DGFather(pos);
    return t;
}
//二维  复杂度(log(n))^2
//小于等于指定位置的元素的翻转(0,0)->(x,y)
void DGDown2(DG_Ran g[][DG_MAXN],DG_Num x ,DG_Num y,DG_Ran av)
{
    while(x >= 0)
    {
        DG_Num tmp = y;
        while (tmp >= 0)
        {
            g[x][tmp] += av;
            tmp = DGBrother(tmp);
        }
        x = DGBrother(x);
    }
}
//获取位置(x,y)的元素翻转次数
DG_Ran DGCUp2(DG_Ran g[][DG_MAXN],DG_Num x ,DG_Num y , DG_Num n)
{
    DG_Ran t = 0;
    while(x < n)
    {
        DG_Num tmp = y;
        while (tmp < n)
        {
            t += g[x][tmp];
            tmp = DGFather(tmp);
        }
        x = DGFather(x);
    }
    return t;
}

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=3398

题目要我们计算1，0的排列方式总数，并且对任意长的字符串，1的数量大于等于0的数量

我们可以把题目转化为从（0，0）点到（m，n）点的方法总数，且路径不经过y=x-1这条直线

/**
 * 线性筛法求素数表
 * 复杂度: O(n)
 */
const long MAXP = 1000000;
long prime[MAXP] = {0},num_prime = 0;
int isNotPrime[MAXP] = {1, 1};
void GetPrime_Init()//初始化调用
{
    for(long i = 2 ; i <  MAXP ; i ++)
    {
        if(! isNotPrime[i])
            prime[num_prime ++]=i;
        for(long j = 0 ; j < num_prime && i * prime[j] <  MAXP ; j ++)
        {
            isNotPrime[i * prime[j]] = 1;
            if( !(i % prime[j]))
                break;
        }
    }
}

线性筛法,即是筛选掉所有合数,留下质数

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=3400

这题就是一道简单的两重三分

首先设e点为从ab上离开的点，f为从cd上进入的点

显然对固定点e,f从距c的距离的长度是一个单调函数或者先减后增的函数，这里可以三分算出最优解。这里是第一个三分

又是我们的OJ

题目链接：

http://www.cn210.com/onlinejudge/problemshow.php?pro_id=92

Description

tancu likes space travels but he is a poor software developer and will never be able to buy his own spacecraft. That is why he is preparing to steal the spacecraft of Petru. There is only one problem - Petru has locked the spacecraft with a sophisticated cryptosystem based on the ID numbers of the stars from the Milky Way Galaxy. For breaking the system Stancu has to check each subset of four stars such that the only common divisor of their numbers is 1. Nasty, isn't it? Fortunately, Stancu has succeeded to limit the number of the interesting stars to N but, any way, the possible subsets of four stars can be too many. Help him to find their number and to decide if there is a chance to break the system.