Brush water problem, brush water problem
portal:http://noi.openjudge.cn/ch0103/
A+b
*c (A+B)
(a+b)/C
04 with remainder Division
05 floating-point values for the calculation of fractions
It's all about seeing the test instructions can read.
#include <iostream>#include<cstring>#include<cstdio>using namespace std; int Main () { int A, b; CIN>>a>>b; cout<<a+b; return 0 ;}
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#include <iostream>#include<cstring>#include<cstdio>using namespace std; int Main () {int a,b,c;cin>>a>>b>>c;cout<< (a+b) *C; return0;}
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#include <iostream>#include<cstring>#include<cstdio>using namespace std; int Main () {int a,b,c;cin>>a>>b>>c;cout<< (a+b)/C; return0;}
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#include <iostream>#include<cstring>#include<cstdio>using namespace std; int Main () {int a,b;cin>>a>>b;cout<<a/b<<' '<<a%b; return 0 ;}
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#include <iostream>#include<cstring>#include<cstdio>using namespace std; int Main () {double a,b;cin>>a>>b;printf ("%.9lf ", A/b) ; return 0 ;}
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http://noi.openjudge.cn/ch0103/06/
Unexpected gain: output% can be used with printf ("percent"); 233 then the output "can use printf (" ""); 233
07 Calculating the value of a polynomial
#include <iostream>#include<cstring>#include<cstdio>using namespace std; int Main () {double a,b,c,d,x;cin>>x>>a>>b>>c>>d;printf ( "%.7lf", a*x*x*x+b*x*x+c*x+D); return 0 ;}
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Actually, just use the POW better?
How about a few days to get a quick polynomial division trial?
08-20
Water problem, a little bit of pit
Direct Sticker Code
#include <iostream>#include<cstring>#include<cstdio>using namespace std; int Main () {double f;cin>>f;printf ("%.5lf" ,5* (f-a)/9) ; return 0 ;}
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#include <iostream> #include <cstring> #include <cstdio>using namespace Std; int Main () { double R;cin >>r;printf ( " %.4lf%.4lf%.4lf " , R*2 , R*2 *3.14159 , R*r*3.14159 ); return 0 ;}
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#include <iostream>#include<cstring>#include<cstdio>using namespace std; int Main () {float r1,r2;cin>>r1>>r2;printf ("%.2LF ",1/(1/r1+1/R2)); return 0 ;}
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#include <iostream>#include<cstring>#include<cstdio>using namespace std; int Main () {double a,b;cin>>a>>b; int c;c=a/b;printf ("%g", a-b*c); return 0 ;}
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#include <iostream>#include<cstring>#include<cstdio>using namespace std; int Main () {double r;cin>>r;printf ("%.2lf" ,3.14*r*r*r*4/3); return 0 ;}
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#include <iostream>#include<cstring>#include<cstdio>using namespace std; int Main () {int n;cin>>n;cout<<n%<< (n/10 )%<<n/; return 0 ;}
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#include <iostream>#include<cstdio>usingnamespace std; int Main () { int v,h,r; CIN>>h>>R; V=h*r*r*3.14; cout<<20000/v+1; return 0 ;}
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#include <iostream>#include<cstdio>#include<cmath>using namespace std; int Main () { double xa,ya,xb,yb; CIN>>xa>>xb>>ya>>YB; printf ("%.3lf", sqrt ((Xa-ya) * (Xa-ya) + (XB-YB) * (xb-YB)); return 0 ;}
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#include <iostream>#include<cstdio>#include<cmath>using namespacestd;intMain () {DoubleA,b,c,r; CIN>>a>>b>>C; if(a+b>c) && (c+b>a) && (a+c>b)) {R= (a+b+c)/2; printf ("%.4LF", sqrt (r* (r-a) * (r-b) * (rc)); } Elsecout<<"Data Error"; return 0;}View Code
#include <iostream>#include<cstdio>#include<cmath>using namespacestd;intMain () {DoubleX1,x2,x3,y1,y2,y3,a,b,c,r; CIN>>x1>>y1>>x2>>y2>>x3>>Y3; A=sqrt ((x1-x2) * (X1-X2) + (y1-y2) * (y1-y2)); b=sqrt ((x2-x3) * (X2-X3) + (y2-y3) * (y2-y3)); C=sqrt ((x1-x3) * (X1-X3) + (y1-y3) * (y1-y3)); R= (a+b+c)/2; printf ("%.2LF", sqrt (r* (r-a) * (r-b) * (rc)); return 0;}View Code
#include <iostream>#include<cstdio>#include<cmath>using namespace std; int Main () { int a1,a2,n; CIN>>a1>>a2>>N; cout<<a1+ (A2-A1) * (n1); return 0 ;}
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#include <iostream>#include<cstdio>#include<cmath>using namespace std; int Main () {longlong a,b;cin>>a>>b;cout<<a*b; return0;}
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#include <iostream>#include<cstdio>#include<cmath>using namespacestd;#definefor (i,j,k) for (int i=j;i<=k;i++)intMain () {Long Longa,b;cin>>a;b=1; For (I,1, a) b=b*2; cout<<b; return 0;}View Code
Unexpected harvest: Helen formula
1.3 Arithmetic expressions and sequential execution of programming basics