//1_2_1: Doubles 求2倍的个数 POJ1552 ZOJ1760 #include #include #include using namespace std; int main() { int i,j,n,cnt,a[20]; scanf("%d",&a[0]); while(a[0] != -1) { n = 1; while(scanf("%d",&a[n

//1_5_10: Number Sequence 求连续数组成的数组串的某位数 POJ1019 ZOJ1410 #include #include #include #include using namespace std; unsigned a[31270],s[31270]; void init() { int i; a[1] = s[1] = 1; for(i = 2;i <

//1_5_8: Ugly Numbers 丑陋数 POJ1338 #include #include #include using namespace std; int min(int a,int b,int c) { int temp = a < b ? a : b; return temp < c ? temp : c; } int main() { int i,l,m,n,cn

//1_5_7: Vertical Histogram 统计字符个数并输出 POJ2136 #include #include #include using namespace std; int main() { int i,j,max,let[26] = {0}; char str[80],letter; for(i = 0;i < 4;i ++) { gets(str);

//1_5_6: The Circumference of the Circle 三点确定圆 POJ2242 ZOJ1090 #include #include #include #include using namespace std; const double pi = 3.141592653589793; int main() { double x1,y1,x2,y2,x3,y3,

//1_5_3: Quicksum 求字符串的校验 POJ3094 ZOJ2812 #include #include #include using namespace std; int main() { int i,sum; char str[260]; while(gets(str),str[0] != '#') { i = sum = 0; while(str[i])

//1_5_1: Sum 求1到N之间的整数和 Ural1068 #include #include #include using namespace std; int main() { int N; scanf("%d",&N); if(N > 0) printf("%d\n",(1 + N) * N / 2); else printf("%d\n",(1 - N) * N /

//1_5_5: Dirchlet's Theorem on Arithmetic Progressions 写第i个素数 POJ3006 #include #include #include #include using namespace std; int isprime(int n) { int i; int sq = (int)sqrt((double)n) + 1; if(

//1_5_4: A Contesting Decision 判定竞赛题目与罚时 POJ1581 ZOJ1764 #include #include #include using namespace std; struct node { char name[20]; int cnt,pun; }team[100]; int main() { int n,i,j,num,time,max

//1_5_2: Specialized Four_Digit Numbers 10,12和16进制的数字和相等 POJ2196 ZOJ2405 #include #include #include using namespace std; int main() { int i,j,sum1,sum2,sum3; for(i = 2992;i < 10000;i ++) { sum

//1_4_2: Humidex 湿热指数，温度和露点之间的关系 POJ3299 #include #include #include #include using namespace std; double get_hum(double tem,double dew) { return tem + 0.5555 * (6.11 * exp(5417.7530 * ((1 / 273.1

//1_3_1: I think I need a houseboat 河水淹没土地 POJ1005 ZOJ1049 #include #include #include using namespace std; const double pi = 3.1415926; int main() { int tem,N,i = 1; double x,y,ans; scanf("%d",&

//1_2_2: Sum of Consecutive Prime Numbers 连续素数之和 POJ2739 #include #include #include #include using namespace std; const int maxn = 2000; int isprime(int n) { int i; if(n == 2) return 1; int sq

//1_1_1: Financial Management 月收入平均值 POJ1004 ZOJ1048 #include #include #include using namespace std; int main() { double ave,sum = 0,mon; int i; for(i = 0;i < 12;i ++) { scanf("%lf",&mon); //

//1_4_1: Hangover 卡片伸向桌子外 POJ1003 ZOJ1045 #include #include #include using namespace std; const int maxn = 300; double len[maxn]; int main() { int i,n = 1; double length; len[n] = 0.5; while(le

//1_5_9: 排列 求下k个排列组合 POJ1833 #include #include #include using namespace std; int a[1030]; void quick_sort(int st,int ed) //快排算法 { int i = st,j = ed,temp; if(st < ed) { temp = a[st]; while(