摘要:POJ 1006 Biorhythms,同百炼2977 生理周期注意边界条件#include <iostream>using namespace std;int mod (int a, int b){ if (a >= 0) return a%b; else return (b-(-a)%b)%b;}int main(void){ int a, b, c, t; int r; for (int i = 1; ; i++) { cin >> a >> b >> c >> t; if (a < 0) ...
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摘要:水题一道//POJ 1005#include <iostream>#include <cmath>using namespace std;int main(){ int caseNum; float x, y; int n; cin >> caseNum; for (int i = 1; i <= caseNum; i++) { cin >> x >> y; n = ceil((x*x+y*y)*3.1415/100); cout << "Property "<<i<<&q
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摘要:求平均数,巨水题#include <iostream>#include <iomanip>using namespace std;int main(void){ float a, sum = 0.0; for(int i = 0; i < 12; i++) { cin >> a; sum +=a; } cout << "$" << sum/12 << setprecision(2) << setiosflags(ios::fixed) << endl; return 0;}
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摘要:POJ 1003 Hangover 水题//POJ 1003#include <iostream>using namespace std;float l[1001];int main(){ float f; l[0] = 0; for (int i = 1; i <= 1000; i++) l[i] = l[i-1]+1.0/(i+1); while (cin>>f, f!=0.00) { for (int i = 1; i <= 1000; i++) if (l[i] > f) { ...
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摘要://空间换时间 POJ 1002 487-3279#include <iostream>#include <cstring>#include <cstdlib>#include <stdio.h>using namespace std;int t[10000000];char change(char c){ if (c < 'Q') return (c-'A')/3+'2'; else return (c-'A'-1)/3+'2';}void add(char*
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摘要:POJ 1000 A+B Promble 巨水#include <iostream>using namespace std;int main(void){ int a,b; cin >> a >> b; cout << a+b << endl; return 0;}
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