hdu5417 Formula method

BCDIV2#52 1001

Test instructions too pit, I thought every time also to calculate machine shut down to meet W value of a variety of take-up operation, every time also to update, think of the scalp tingling, the results of a long time directly abandoned BC, can not go on.

Results just when the problem found in clarification in the 1001 example of the resolution, just know in the moment of closing is allowed to pop the ball, instead of popping the ball. It's a mistake to understand.

The rest is an understanding of mathematical calculations. My math is weak .... It's all holes.

The formula in the code, for me really hard to think Ah ....

I think we must pay attention to the meaning of TT, otherwise it's definitely a pit.

1#include <cstdio>2#include <iostream>3#include <cstring>4#include <cstdlib>5 6 using namespacestd;7 intX,y,w,n,num,res,tt,ss;8 9 intMain ()Ten { One      while(cin>>x>>y>>w>>N) { A         if(x<W) { -Res= (n1) * (x+y); -         } the         Else{ -num=1+x/w;//you can eject num balls in x time -Tt=n%num;//tt=0 indicates that the time of the SS X is enough to eject n balls, but there may be surplus; -             //tt! =0 indicates the time required to add tt-1 W outside the time of the SS X +Ss=n/num;//takes a few x long hours -             //cout<< "num" <<num<< "ss" <<ss<< "tt" <<tt<<endl; +             if(tt!=0) ARes=ss* (x+y) + (tt-1+num)%num*W; at             Else  -Res= (ss-1) * (x+y) + (tt-1+num)%num*w;// -         } -cout<<res<<Endl; -     } -     return 0; in}

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Be stunned by the code of XXX .... A similar way of solving a problem, but with a completely different wording

1#include <cstdio>2#include <iostream>3#include <cstring>4#include <cstdlib>5 6 using namespacestd;7 8 intMain ()9 {Ten     intx,y,w,n,res=0, tmp,mul,ans=0;  One      while(cin>>x>>y>>w>>N) { Atmp=x/w+1;//x time the number of pop-up balls -         //cout<<tmp<< ""; -Mul= (n1)/tmp;//How many times x time the         //cout<<mul<< ""; -Ans= (x+y) *mul;//ans Initialization -         //cout<<ans<< ""; -n-=mul*tmp;//N Value Update +         //cout<<n<<endl; -n--; +          while(n--){ Aans+=W; at         } -cout<<ans<<Endl; -     } -     return 0; -}

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Another way to imitate the web search

1#include <stdio.h>2#include <string.h>3#include <iostream>4#include <algorithm>5#include <vector>6#include <queue>7#include <stack>8#include <Set>9#include <map>Ten#include <string> One#include <math.h> A#include <stdlib.h> -#include <time.h> - using namespacestd; the intMain () - { -         intX,Y,W,N,ANS,I,TP; -      while(~SCANF ("%d%d%d%d",&x,&y,&w,&N)) +     { -I=1; +ans=tp=0; A          while(i<N) at         { -             if(ans+w<tp+x) -             { -ans+=W; -i++; -             } in             Else if(ans+w==tp+x) -             { toi++; +tp+=x; -ans=TP; the                 if(i==n) Break; *i++; $tp+=y;Panax Notoginsengans=TP; -             } the             Else if(ans+w>tp+x) +             { Ai++; thetp=tp+x+y; +ans=TP; -             } $         } $printf"%d\n", ans); -     } -     return 0; the}

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The first time to knock the simulation, almost crying, fortunately, I AC

1#include <cstdio>2#include <iostream>3#include <cstring>4#include <cstdlib>5 6 using namespacestd;7 8 intMain ()9 {Ten     intx,y,w,n,tmp,mul,ans=0;  One      while(cin>>x>>y>>w>>N) { A         intres=0; -             if(w>x) { -Res= (n1) * (x+y); the             } -             Else if(w==x) { -                  while(n!=0){ -n--; +                     //cout<< "n" <<n<<endl<<res<<endl; -                     if(n>1) +                     { Ares+=w+y; atn--;  -                     } -                     Else if(n==1) -                     { -res+=W; -n--; in}//cout<< "res" <<res<<endl; -                 } to             } +             Else { -                  while(n!=0) the                 { *n--; $                     intans=0;Panax Notoginseng                      while(ans+w<=x&&n!=0){ -ans+=W; then--; +                         //cout<< "ans" <<ans<<endl; A                     } the                     if(n!=0) +                     { -res+=x+y; $                     } $                     Elseres+=ans; -                     //cout<< "res" <<res; -                 } the             } -cout<<res<<Endl;Wuyi     } the     return 0; -}

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hdu5417 Formula method