题目  UVA 100 - The 3n + 1 problem解题思路：题目大意奇数就乘3加1，偶数就除以2，直到这个数变成1，我WA了几次，后来发现i，j的值被我改了，还有一个要注意的是输入的第一个数不一定小于第二个数，数据用int就OK了，时间卡的不是很死，所以不用打表做。#includeint f(int i){ int n=0; while(i!=1)

#include#includeint B1,B2,B3,G1,G2,G3,C1,C2,C3;char str[6][4]={"BCG","BGC","CBG","CGB","GBC","GCB"};int an[6]={0};int main(){ while(scanf("%d%d%d%d%d%d%d%d%d",&B1,&G1,&C1,&B2,&G2,&C2,&B3,&G3

#include#include#includeusing namespace std;char s[200];int T;stackS;int main(){ int i,len; scanf("%d",&T); getchar(); while(T--) { gets(s); while(!S.empty(

#include#include#includeint T,n,day;int an[10000];int flag[10000];int main(){ scanf("%d",&T); while(T--) { //int n; memset(an,0,sizeof(an)); memset(flag,0,si

用循环双向链表模拟#include#include#includestruct s{ int num; struct s *qian; struct s *hou; int flag;};int n,k,m;int main(){ while(scanf("%d%d%d",&n,&k,&m)!=EOF) { i

模拟，我也不知道开始错在那里，重新敲了一遍就过了#include#includestruct s{ char c[60][5]; int s;}card[60];void deal(int a,int b){ int x=card[a].s; int y=card[b].s; strcpy(card[a].c[x+1],card[b

#include#includeint an[105];int main(){ int T; scanf("%d",&T); while(T--) { int n; scanf("%d",&n); int sum=0; char s1[10],s2[10]; int t;

#include#includeint main(){ int T; scanf("%d",&T); while(T--) { int x,y,z; scanf("%d%d%d",&x,&y,&z); //int t=(x+y)/3; printf("%d\n",(2*x-y)*z/(x+y))

#include#include#include#includedouble H,U,D,F;int main(){ while(scanf("%lf%lf%lf%lf",&H,&U,&D,&F)!=EOF&&H) { int flag=0; int day=0; double climbH=0; do

注意0就好#include#include#includeint main(){ int T,n; scanf("%d",&T); while(T--) { scanf("%d",&n); n=(int)abs(n); int i,s=0; if(n==0) {

今天水了一题直接上代码，没什么好说的 #include#include#include#includechar s[10000];int main(){ long long sum; while(scanf("%s",s)!=EOF) { if(!strcmp("0",s)) break;

自己看代码，水题  #includeint main(){ int N,M; while(scanf("%d%d",&N,&M)!=EOF) { printf("%d\n",(N-1)+(M-1)*N); } return 0;}

解题思路：推公式  (n+1)a1=n*a0+an+1-2*[(n-0)*c1+......+(n-n+1)*cn]  #includeint main(){ int T; scanf("%d",&T); while(T--) { int n,i; double a0,an,c,cn,a1; scanf

解题思路：水题目一枚，直接代码吧，注意的是最外面边界不用考虑the border cells do not need to be controlled, because Nessy cannot hide there (she is too big). #include#include#includeint main(){ int n,m,t; scanf(

题意我理解了很久，英语不好的伤不起，把英文看懂就知道写了，纯模拟题，不难，题意明白就好，纯C写的 #include#include#define MAX 100char s[MAX];char s1[MAX];int pang(char c);int f1();void fun();int pang(char c){ int i,flag=0; i

其实就是3n+1问题，多了一个限制而已，运算时数据会超过int，开始用int64，发现uva用不了，改成longlong就可以了AC #includevoid fun(long long A,long long L){ int flag=1; while(A!=1) { if(A%2!=0) A=A*3+1;

#includevoid fun(int z,int p){ int i,j,k; for(i=0;i<p;i++) { for(j=1;j<=z;j++) { for(k=1;k<=j;k++) { printf("%d",j); }

题目  UVA 10189 Minesweeper解题思路：扫雷问题  找‘.’的八个方向的地雷数，很简单，需要注意的有注释，看代码#include#include#define Max 110char s[Max][Max];int main(){ int n,m,i,j,flag=0,t,k,l; while(scanf("%d%d",&n,&m)!=EOF) {

此题看了解题报告，觉得他的代码写的非常好，果断转他的了#include#includeint main(){ int T,x,y,z,a; scanf("%d",&T); while (T--) { scanf("%d%d",&x,&y); if (y==x) printf("0"); else