把一张一元的钞票,换成一分,二分,和五分的硬币,每种至少五枚,问有多少种方案?(求编写一个C语言程 把一张一元钞票,换成一分、二分和五分硬币,每种至少5枚,问有...

[\u6709\u7a0b\u5e8f\u4e0d\u592a\u61c2\u6c42\u8be6\u89e3]\u628a\u4e00\u5f20\u4e00\u5143\u7684\u949e\u7968\uff0c\u6362\u6210\u4e00\u5206\uff0c\u4e8c\u5206\uff0c\u548c\u4e94\u5206\u7684\u786c\u5e01\uff0c\u6bcf\u79cd\u81f3\u5c11\u4e94\u679a\uff0c\u95ee\u6709\u591a\u5c11\u79cd\u65b9\u6848

\u56e0\u4e3a\u9898\u76ee\u4e2d\uff0c\u6bcf\u79cd\u6570\u989d\u90fd\u81f3\u5c11\u89815\u679a\uff0c\u6240\u4ee5\u521d\u59cb\u503c\u4e3a5
\u7136\u540e\u4e5f\u5c31\u662f\u8bf4\u521d\u59cb\u81f3\u5c112\u5206\u7684\u67095\u679a\uff0c5\u5206\u7684\u4e5f\u67095\u679a\uff0c\u52a0\u8d77\u6765\u503c35\u5206\uff0c\u6240\u4ee5\u57281\u5206\u7684\u6700\u5927\u8303\u56f4\u4e2d\uff0c\u8981\u51cf\u53bb\u90a335\uff0c\u526965
\u5176\u5b83\u7684\u540c\u7406
\u521d\u59cb\u4e2d1\u5206\u76845\u679a\uff0c5\u5206\u76845\u679a\uff0c\u52a0\u8d77\u6765\u503c30\u5206\uff0c\u6240\u4ee52\u5206\u7684\u6700\u5927\u8303\u56f4\u503c\u4e3a70\u5206\uff0c\u4e5f\u5c31\u662f\u670935\u679a2\u5206\u7684
\u521d\u59cb\u4e2d1\u5206\u76845\u679a\uff0c2\u5206\u76845\u679a\uff0c\u52a0\u8d77\u6765\u503c15\u5206\uff0c\u6240\u4ee55\u5206\u7684\u6700\u5927\u8303\u56f4\u503c\u4e3a85\u5206\uff0c\u4e5f\u5c31\u662f\u670917\u679a5\u5206\u7684

\u56de\u7b54\u5982\u679c\u6ee1\u610f\uff0c\u671b\u91c7\u7eb3

\u81f3\u5c115\u679a\u5c31\u4e00\u5171\u82b1\u4e864\u89d2\u3002\u4e5f\u5c31\u662f\u5269\u4e0b6\u89d2\u7684\u5206\u914d\u95ee\u9898\uff0c5\u5206\u768412\u4e2a1\u79cd\uff1b5\u520611\u4e2a\uff0c\u67093\u79cd\uff1b5\u5206\u768410\u4e2a\uff0c\u67096\u79cd\uff1b5\u52069\u4e2a\uff0c8\u79cd\uff1b5\u52068\u4e2a\uff0c11\u79cd\uff1b5\u52067\u4e2a13\u79cd\uff1b5\u52066\u4e2a\uff0c15\u79cd\uff1b5\u52065\u4e2a\uff0c18\u79cd\uff1b5\u52064\uff0c20\u79cd\uff1b5\u52063\u4e2a\uff0c23\u79cd\uff1b5\u52062\u4e2a\uff0c25\u79cd\uff1b5\u52061\u4e2a28\u79cd\uff1b5\u52060\u4e2a\uff0c30\u79cd\uff1b\u4e00\u5171\u5c31\u662f201\u79cd\u3002

/*把一张一元的钞票,换成一分,二分,和五分的硬币,每种至少五枚,问有多少种方案*/
#include <stdio.h>
 
int main() {
    int n1, n2, n5;
    for (n1 = 5; n1 <= 65; n1++)        //因为2和5至少5枚,所以1至多有65
        for (n2 = 5; n2 <= 35; n2++)    //同理1和5至少5枚,所以2至多有35
            for (n5 = 5; n5 <= 17; n5++)//同上
                if (n1 * 1+ n2 * 2 + n5 * 5 == 100)
                    printf("%3d * 1 + %3d * 2 + %3d * 5 = 100
", n1, n2, n5);
    return 0;
}


#include <stdio.h>

int main() {
int n1, n2, n5;
for (n1 = 5; n1 <= 65; n1++)
for (n2 = 5; n2 <= 35; n2++)
for (n5 = 5; n5 <= 17; n5++)
if (n1 + n2 * 2 + n5 * 5 == 100)
printf("%3d + %3d * 2 + %3d * 5 = 100
", n1, n2, n5);
return 0;
}


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