codes often WA
枚举:
1.完美立方
#include<iostream>
#include <cstdio>
using namespace std;
int main()
{
int N;
scanf("%d",&N);
for (int a = 2; a <= N; ++a)
for(int b = 2;b< a;++b)
for(int c = b;c < a;++c)
for(int d = c;d < a;++d)
if(a * a * a == b *b *b + c*c*c+d*d*d)
printf("Cube = %d,Triple =(%d,%d,%d)\n",a,b,c,d);
return 0;
}
2.
生理周期
原题如下:
Time Limit: 1000MS | Memory Limit: 10000K | |
Total Submissions: 110700 | Accepted: 34443 |
Description
Since the three cycles have different periods, the peaks of the three cycles generally occur at different times. We would like to determine when a triple peak occurs (the peaks of all three cycles occur in the same day) for any person. For each cycle, you will be given the number of days from the beginning of the current year at which one of its peaks (not necessarily the first) occurs. You will also be given a date expressed as the number of days from the beginning of the current year. You task is to determine the number of days from the given date to the next triple peak. The given date is not counted. For example, if the given date is 10 and the next triple peak occurs on day 12, the answer is 2, not 3. If a triple peak occurs on the given date, you should give the number of days to the next occurrence of a triple peak.
Input
Output
Case 1: the next triple peak occurs in 1234 days.
Use the plural form ``days'' even if the answer is 1.
Sample Input
0 0 0 0
0 0 0 100
5 20 34 325
4 5 6 7
283 102 23 320
203 301 203 40
-1 -1 -1 -1
Sample Output
Case 1: the next triple peak occurs in 21252 days.
Case 2: the next triple peak occurs in 21152 days.
Case 3: the next triple peak occurs in 19575 days.
Case 4: the next triple peak occurs in 16994 days.
Case 5: the next triple peak occurs in 8910 days.
Case 6: the next triple peak occurs in 10789 days.
Source
#include <iostream>
#include<cstdio>
using namespace std;
#define N 21252;
int main()
{
int p,e,i,k,d,caseNO = 0;
while(cin >>p>> e>>i >>d && p != -1)
{
++caseNO;
for(k = d+1 ;(k-p)%23;++k);
for(;(k-e)%28; k+= 23);
for (;(k-i)%33;k+=23*28);
cout << "caseNO" << ": the next triple peak occurs in " << k-d<<endl;
}
return 0;
}
3.称硬币
#include <iostream>
#include <cstring>
using namespace std;
char Left[3][7];
char Right[3][7];
char result[3][7];
bool isFake(char c,bool light);
int main()
{
int t;
cin >> t;
while(t--)
{
for(int i = 0; i < 3; i++)
cin >> Left[i] >> Right[i] >> result[i];
for(char c = 'A';c <= 'L';c++)
{
if(isFake(c,true))
{
cout << c << "is the counterfeit coin and it is light.\n";
break;
}
else if (isFake(c,false))
{
cout << c << "is the counterfeit coin and it is heavy.\n";
break;
}
}
}
return 0;
}
bool isFake(char c,bool light)
{
for (int i=1;i<=3;i++)
resul
if ()
return true;
else
return false;
}
4,熄灯问题
#include<memory>
#include<string>
#include<cstring>
#include <iostream>
using namespace std;
char oriLights[5];
char lights[5];
char result[5];
int GetBit(char c,int i)
{
return (c >> i) & 1;
}
void SetBit(char & c, int i ,int v)//c 的第i位变位v
{
if(v == 1){
c |= ( 1 << i);//c 或 后移i个位置
}
else // v == 0
c &= ~(1 << i); //c的第i位为1 ; 取反(第i位成0);与(第i位成0)
}
void FlipBit(char &c, int i)
{
c ^= ( 1 << i);
}
void OutPutResult(int t,char result[])
{
cout
}
---------------------------------------分隔符-------------------------------------------------------------------
while(~scanf("%d%d",&m,&n))等同于while (scanf("%d%d",&m,&n)!=EOF)
---------------------------------------分隔符-------------------------------------------------------------------
#include <stdio.h>
#define maxsize 32575
typedef int SElemType;
typedef struct stack{
SElemType *base,*top;
int stacksize;
}stack;
int Initstack(stack &S){ &s S *S &S
S.base = new SElemType[maxsize];
if(!S.base)
return -1;
S.base = S.top;
S.stacksize = maxsize;
return 0;
}
int push(stack &S,SElemType e){
if(S.top -S.base == S.stacksize)
return -1;
*(S.top++) = e;
return e;
}
int pop(stack &S,SElemType &e){
if(S.top == S.base)
return -1;
e = *--S.top;
return 1;
}
int stackEmpty(stack S){
if(S.base == S.top)
return -1;
return 0;
}
//十进制转换为八进制
int main(int a){
stack S;
SElemType e;
Initstack(S);
stackEmpty(S);
scanf("%d",&a);
while(a){
push(S,a%8);
a = a/8;
}
while(!stackEmpty(S)){
pop(S,e);
printf("%d",e);
}
return 0;
}
---------------------------------------分隔符-------------------------------------------------------------------
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