N

<br />//简单模拟，重点掌握string cin cout#include <iostream>#include <string>using namespace std;string calc(string str){ int i, j = 0; int num[10] = {0}; char inv[81]; for(i = 0; i < str.length(); i++) //c++ ! { num[str[i]-'0']++; }

<br />//典型的活动会场安排问题//c中竟然不能用bool？#include <stdio.h>#include <stdlib.h>#define csMax 5001struct data{ int length; int weight; int flag;};struct data sticks[csMax]; //c中struct不能省，c++中可以！int cmp(const void *a, const void *b){ struc

<br />}//唯一要主义的是字符%的输出 printf("%%");#include <stdio.h>int main(){ int i; int nCase, nNum, nGrade[1001]; int temp, average; while (1==scanf("%d", &nCase)) { while(nCase--) { temp=0; scanf("%d", &nNum); for(i=0; i<nNum

<br />//简单模拟#include <stdio.h>#define csMax 101int main(){ int i, j, k; int nCommands; int x[csMax], y[csMax], L[csMax]; int flag[csMax][csMax]={0}; char command[csMax][6]; int a1, b1, count, temp; while (1==scanf("%d", &nC

 qsort的运行，具体看MSDN    一个是二位数组的qsort用法：1intcompare( constvoid*arg1, constvoid*arg2 ) 2{ 3   returnstrcmp((char*)arg1, (char*)arg2 ); 4} 5intarr[n][11]; 6qsort(arr, n, sizeof(arr[0]), compare);#include #include #include #define csMax 100000+1char nN

/*T>G>C>ACCCGGGGGGA 9AACATGAAGG 3+5+2=10GATCAGATTT 4+4+2+1=11ATCGATGCAT 6+2+2+3+2+1=16TTTTGGCCAA 24+8+4=36TTTGGCCAAA 21+10+6=37*/#include #include #include struct sortNum{ char str[51]; int count;};struct sortNum t[1

<br />//2010-11-15#include <stdio.h>#include <stdlib.h>struct vote{ int firstRound; int secondRound; int ID;};struct vote arr[50001]; int cmp(const void *a, const void *b){//必须这样写！用“->”错！ 明白此处 “.” 和“ ->” 的区别！！！！ return (*(s

<br />思路一、排序暴力<br />//a1x1 3+ a2x2 3+ a3x3 3+ a4x4 3+ a5x5 3=0 <br />//拆成a3x33+ a4x43+ a5x53 = -（a1x1 3+ a2x2 3）<br />#include <stdio.h><br />#include <stdlib.h><br /> <br />int cmp(const void *a, const void *b)<br />{<br />       return *(int *)a - *(int

<br />//快排 充分利用qsort()函数！#include <stdio.h>#include <string.h>#include <stdlib.h>int cmp_w(const void *a, const void *b){ return strcmp((char *)a, (char *)b);}int cmp_str(const void *a,const void *b){ return *(char *)a-*(char *)b;

<br />考察基本功<br />1、getchar <br />语法: <br />  #include <stdio.h><br />  int getchar( void );<br />getchar()函数从STDIN(标准输入)获取并返回下一个字符,如果到达文件尾返回EOF.<br />2、gets <br />语法: <br />  #include <stdio.h><br />  char *gets( char *str );<br />gets()函数从STDIN(标准输入)读取字符并

V = (|A(1) – B(1)| + |A(2) – B(2)| + ... + |A(N) – B(N)|) + (|B(1) – B(2)| + |B(2) – B(3)| + ... +|B(N-1) – B(N)|)<br />将公式简单变形就可以推出b[0]=a[0]和b[n-1]为区间[a[n-1],b[n-2](不知道a[n-1]与b[n-2]大小关系)内的任意一个数或者b[n-1]=a[n-1]和b[0]为区间[a[0],b[1]]区间内的任意一个数,而考虑b[i],即要使 |b[i]-

<br />学习点 PI=acos（-1..0） sin(x), cos(x), x是弧度！<br />#include <stdio.h>#include <math.h>//#define PI 3.1415926535#define PI acos(-1.0) //PI的表示还是值得学习的 int main(){ int nCase, nCaseNum; double R, r; int n; nCaseNum=0; while (1==scanf("%d"

<br />#include <stdio.h>#include <string.h>int main(){ int i; int n, m; int nCard, flag[1001]; int caseNum = 1; int count, res, temp; while (scanf("%d %d", &n, &m) != EOF, n, m) { memset(flag,0, sizeof(flag)); for(i = 0; i

输入n个二元组(xi,yi)，统计有多少个二元组(xk,yk)满足：对于任意xi yk；对于任意yi xk ( n

#include using namespace std;const int maxn = 2500;int nFactorial[maxn];int funcSolve(int n);int main(){ int num; while(1 == scanf("%d", &num)) { int result = funcSolve(num); printf("%5d -> %d/n", num, result); //输出要注意！ }

争取做完上面所有的题目。

<br />模拟得我吐血！<br />有空理理思路，重新做一遍！

<br />#include <iostream>#include <algorithm>#include <math.h>#define csMax 1001using namespace std;struct LocationData{ double nFrom; double nTo;};LocationData dd[csMax];int nCaseNum, nIslandNum;double d;double nX[csMax], n

<br />/*POJ 2014 Flow Layout 模拟*/#include <iostream>using namespace std;int main(){ int maxGiven; int width_in, height_in; int width_curr, height_curr; int width_max, height_max; while(1 == scanf("%d", &maxGiven)) { if(maxGiven

<br />/*推导公式：(p+q+10)/7 */#include <iostream>using namespace std;int main(){ int nCase; int p, q; double result; scanf("%d", &nCase); while(nCase--) { scanf("%d %d", &p, &q); result= (p + q + 10) / 7.0; printf("%.3lf/n",

<br />/* %g(%G) 浮点数不显无意义的零"0" */#include <iostream>#include <algorithm>using namespace std;void funcInput();void funcDealData();bool cmp(const int &a, const int &b);void funcOutput();int num[7];int judge, sum;int main(){

<br />#include <iostream>#include <math.h>using namespace std;const int csMax = 1299709;bool isPrim(int a);int main(){ int num, maxGap; while(1 == scanf("%d", &num), num) { if(isPrim(num)) { maxGap = 0;; } else

<br />/*注意&的使用 num & (num-1) 判断num是否为2的k次方 证明看：http://hi.baidu.com/findthegateopen/blog/item/03e5802e19f1ea301e3089f9.html*/#include <iostream>using namespace std;int main(){ int num; while(1 == scanf("%d", &num)) { printf((num &

<br />/*题意理解+细节+精度*/#include <iostream>#include <cmath>using namespace std;int main(){ int b, w; int size; while(2 == scanf("%d %d", &b, &w)) { if(b == 0 && w == 0 ) { printf("Impossible/n"); } else { if(w ==

<br />思路：见代码注释。<br /> <br />#include <iostream>#include <cmath>using namespace std;int nPersons, suriverNum;int funcInput();void funcWorkout();void funcOutput();int main(){ while(funcInput()) { funcWorkout(); funcOutput(

<br />1、c++中字符串的操作头文件 #include <string><br />2、getline<br />3、find<br /> <br />#include <iostream>#include <string> //c++中字符串操作的头文件！using namespace std;int main(){ string str; while(getline(cin, str)) //getlin 函数 { int start = str

<br />2010-12-10 15:44:04<br /> <br /> <br /> <br />/*功能： 函数返回参数arg的绝对值。 1、fabs 语法: #include <math.h> double fabs( double arg ); 2、abs 语法: #include <stdlib.h> int abs( int num );*/#include <iostream>using namespace s

a -> 1<br /> b -> 2<br /> .<br /> .<br /> z -> 26<br /> ab -> 27<br /> ac -> 28<br /> .<br /> .<br /> az -> 51<br /> bc -> 52<br /> .<br /> .<br /> vwxyz -> 83681/*理解是关键，数论题*/#include <

1、qsort函数的应用int cmp(const void *a, const void *b){ struct data *c = (struct data *)a; //注意struct不能省！ struct data *d = (struct data *)b; //！ if((c->length) == (d->length))  return c->weight - d->weight; else  return c->length - d->length;}//典型的活动会场安排问题//c