HDU2015

HDU2015偶数求和

偶数求和TimeLimit:2000/1000MS(Java/Others)MemoryLimit:65536/32768K(Java/Others)TotalSubmission(s):69758AcceptedSubmission(s):29740ProblemDescription有一个长度为n(nintmain(){inti,j,a,n,m,sum;while(scanf("%d%d",&
AC_Dreameng·2020-08-21 14:58

关于已解决HDU题解链接

HDU2015求偶数和HDU1040AsEasyAsA+BHDU1062TextReverseHDU1094A+BforInput-OutputPractice(VI)HDU1228A+BHDU1720A
m0_38056893·2020-07-28 03:10

hdu2015偶数求和

ProblemDescription有一个长度为n(nusingnamespacestd;intmain(){intn,m;while(cin>>n>>m){intjudge=0,sheng,zu;intave[100]={0};//数组清零还有函数可以处理for(inti=0;i1){coutintmain(void){intn,m;while(scanf("%d%d",&n,&m)!=EOF)
多安_0be4·2019-05-21 11:06

HDU2015 偶数求和

问题链接：HDU2015偶数求和。入门训练题，用C语言编写程序。这个问题是对n项的等差数列（公差为2），每m项计算一个平均值输出。
tigerisland45·2016-06-27 08:00

HDU2015偶数求和

偶数求和TimeLimit:2000/1000MS(Java/Others)    MemoryLimit:65536/32768K(Java/Others)TotalSubmission(s):69758    AcceptedSubmission(s):29740ProblemDescription有一个长度为n(n intmain() { inti,j,a,n,m,sum; while(sc
hurmishine·2016-03-01 14:00

hdu2015――偶数求和

原题： ProblemDescription 有一个长度为n(nintmain(){inti,n,m,b,c;while(scanf("%d%d",&n,&m)!=EOF){b=2;c=0;for(i=0;i<n/m;i++){printf(c++?"%d":"%d",b+m-1);b+=m*2;}printf(n%m?"%d\n":"\n",b+n%m-1);}return0;} 
bingsanchun·2013-03-05 21:00

HDU2015

偶数求和 Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 17853    Accepted Submission(s): 7718 Problem Descrip
Dev|il·2011-08-09 14:00
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